I was helping my brother with his Algebra II. He was beginning a chapter on sequences and series. The textbook showed:
1, 4, 9, 16, 25, ...
Obviously, it's 1 squared, then 2 squared, etc.
But he noticed a different pattern: start with 1. Add 3, then add 5, then add 7, and so on. The difference between one number and the next increases by 2 each time.
4 - 1 = 3
9 - 4 = 5
16 - 9 = 7
25 - 16 = 9
That's cool! But we're both curious - why does this happen?
You might find a few more interesting things about squares.
* A square, sometimes called a perfect square, is the result of multiplying a number by itself as in N = nxn = n^2.
* The perfect squares are the squares of the counting numbers, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, etc. which produce 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, etc.
* The last digit in the square of a number must be one of the following: 0, 1, 4, 5, 6, or 9. Note that there are numbers that end in these digits that are not squares but to be a square, they must end in one of these digits.
* If the last digit of a number is 0, its square ends in 00 and the preceding digits form a square.
* If the last digit of a number is 1 or 9, its square ends in 1 and the number formed by the preceding digits must be divisible by 4.
* If the last digit of a number is 2 or 8, its square ends in 4 and the preceding digit must be even.
* If the last digit of a nuber is 3 or 7, its square ends in 9 and the number formed by the preceding digits must be divisible by 4.
* If the last digit of a number is 4 or 6, its square ends in 6 and the preceding digit must be odd.
* If the last digit od a number is 5, its square ends in 25 and the preceding digits must be 0, 2, 06 or 56.
* The last two digits of a 3 or more digit square number must be one of the following: 00, 01, 04, 09, 16, 21, 24,
25, 29, 36, 41, 44, 49, 56, 61, 64, 69, 76, 81, 84, 89, or 96.
* The sum of the digits in a square MUST add up to 1, 4, 7, or 9.
* The perfect squares are simply the successive sum of the odd numbers.
0 + 1 = 1
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
1 + 3 + 5 + 7 + 9 = 25, etc.
* The perfect squares can also be expressed by the sum of 1+1+2+2+3+3+....(n-1)+(n-1)+n. For example, 5^2 = 1+1+2+2+3+3+4+4+5 = 25.
* The nth square can be determined from the previous two from n^2 = 2(n - 1)^2 - (n - 2)^2 + 2. For example, 6^2 = 25 + 25 - 16 + 2 = 36.
* The nth square is equal to n^2.
n....1....2....3....4....5....6....7....8....9....10
Sq..1...4....9....16..25..36..49..64...81...100
* The nth non-square is n + sqrt(n) r.o. (r.o. = rounded off)
n....................1....2....3....4....5....6....7....8....9....10....11....12....13....14....15
Non-square.....2....3....4....6....7....8...10..11...12...13....14....15....17....18....19
6th non-square = 6 + 2[.449] = 8; 12th non-square = 12 + 3[.464] = 15.
* The perfect squares are also the sum of consecutive triangular numbers, Tn = n(n + 1)/2 = 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, etc.
1^2 = 0 + 1 = 1
2^2 = 1 + 3 = 1 + 3
3^2 = 1 + 3 + 5 = 3 + 6
4^2 = 1 + 3 + 5 + 7 = 6 + 10
5^2 = 1 + 3 + 5 + 7 + 9 = 10 + 15 - 1, 3, 6, 10, 15, etc., being the sequential triangular numbers.
* All even squares are congruent to 0 modulo 4 meanig that all even squares minus 0 are divisible by 4. More simplistically, all even squares are evenly divisible by 4.
* All odd squares are congruent to 1 modulo 8 meaning that all odd squares minus 1 are evenly divisible by 8. Another way of staing this is that the square of an odd number is always of the form 8n + 1.
* The square of a number is either divisible by 4 or leaves a remainder of 1 when divided by 4.
* The square of an odd number is always of the form 8n + 1.
* The sum of the first n squares, i.e., 1^2 + 2^2 + 3^2 + 4^2 + .....+ n^2 = n(n + 1)(2n + 1)/6.
* An easy way to determine the square of N, s to find two numbers whose mean is N from which N^2 = (N+1)(N-1) + 1. For example, for N = 25, 25 is the mean of 24 and 26. Therefore, the square of 25 is 24x26 + 1 = 625.
* An easy way to derive the squares: Having x^2, the value of (x + 1)^2 can be expressed by (x + 1)^2 = x^2 +
(2x + 1). What does this mean? If you know the square of a particular number, say 30^2 = 900, then the square
of 31 is 31^2 = 30^2 + (2(30) + 1) = 900 + 60 + 1 = 961. Similarly, the square of (x - 1) is defined by (x - 1)^2 =
x^2 -2x + 1 and for 29^2 we have 29^2 = 900 - 60 + 1 = 841. We also know that (x + 2)^2 = x^2 + 4x + 4 and (x
+ 3)^2 = x^2 + 6x + 9. This can be easily written as (x + n)^2 = x^2 + (2xn + n^2) and in the case of (x - n)^2 =
x^2 -(2xn + n^2). Therefore, knowing the simple squares such as 20^2, 30^2, 70^2 90^2, 120^2, etc., it is an
easy computation to arrive at the adjacent 5 squares.
*--Did you ever notice anything odd about the series of squares? 1-4-9-16-25-36-49-64-81-100-121-144-169-etc.
Squares - 1 - 4 - 9 - 16 - 25 - 36 - 49 - 64 - 81 - 100 - 121 - 144 - 169
Differences - 3 5 7 9 11 13 15 17 19 21 23 25 and so on ad infinatum. If you happen to
know two adjacent squares, such as 15^2 = 225 and 16^2 = 256, you immediately know the next squares on
either side of the two known ones. If you wanted to know the square of 23, you could use the expression (x +
n)^2 = x^2 + (2n + n^2) = (20 + 3)^2 = 20^2 + (2(20)3 + 9) = 400 + 129 = 529 or the same expression to find (x
+ 1)^2 = 441, obtain the difference from 441 and 400, and apply it to (x + 3)^2 = 400 + 41 + 43 + 45 = 529.
* Having one square, x, we know that the next square is (x + 1)^2 = x^2 + (2x + 1).
* The square of any number ending in 5 is obtained from x(x + 1) and placing a 25 after the result, e.g., 25^2 =
2(2 + 1) plus 25 = 6 + 25 = 625.
* If an integer N>1 is not a perfect square, then sqrtN is irrational, i.e., sqrtN cannot be expressed as a/b, where a and b are integers.
* Any integer which is a ratio of squares is, itself, a square.
* If both a and b are perfect squares, then a(b) is a perfect square.
* There exists sets of 4 integers such that the sum of all four of them and the sum of each pair of them are squares.
Verify with 386, 2114, 3970 and 10,430.
* Lagrange's Theorem - Any whole number can be decomposed into sums of up to four squares.
Example - 97 = 81 + 16 = 64 + 25 + 4 + 4.
* 1/10th of the first 100 integers are perfect squares. 1/100th of the first 10,000 numbers are perfect squares. 1/1000th of the first 1,000,000 numbers are perfect squares.
* The product of three consecutive integers is never a square.
* An easy way of determining the square of a number is to identify two numbers that are the mean of the number in question, multiply them together and add the square of the distance from the mean.
Example: 24^2 = (23x25) + 1^2 = 576 or 16^2 = (14x18) + 2^2 = 196.
* Perhaps obvious:
The squares of odd square numbers are odd and the squares of even square numbers are even.
Conversely, the square roots of odd numbers are odd and the square roots of even nubers are even.
* While not very useful in calculating squares, it is worth noting that a square is the sum of 1+1+2+2+3+3+...(n-1) + (n-1)+n.
Example: 5^2 = 1+1+2+2+3+3+4+4+5 = 25.
* Sums of Squares
1--Any prime number of the form 4n + 1 can be represented as the sum of two integral squares.
2--Any product whose factors are 2 and primes of the form 4n + 1 can be represented as the sum of two
integral squares.
3--As derived by Diophantus, the product of two numbers, each of which is the sum of two squares, can be
expressed as the sum of two squares in at least two ways from
...................(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2
...................(a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2
....................2(a^2 + b^2) = (a + b)^2 + (a - b)^2
(one can represent the product of any two numbers that are sums of two squares as the sum of two other
squares, and even in two different ways. An example will illustrate.
Show how the numbers 481, 685, and 1690, can be written as the sum of two integral
squares.
Represent the number 481 as the sum of two integral squares.
.............(481 - 1)/4 = 120 so there are answers.
1--481 = 13 x 37 both prime factors being of the form 4x + 1.
2-- 13 = 3^2 + 2^2 and 37 = 6^2 + 1^2
............a.......b...........................c.......d
3--Using the identity and the a, b, c, d designations shown:
For (a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2
..........................13 x 37 = (3x6 + 2x1)^2 + (3x1 - 2x6)^2
......................................= (18 + 2)^2 + ( 3 - 12)^2
.......................................= (20)^2 + (9)^2
Similarly, for (a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2, .......................................13 x 37 = (3x6 - 2x1)^2 + (3x1 + 2x6)^2
...................................................= (18 - 2)^2 + ( 3 + 12)^2
...................................................= (16)^2 + (15)^2
Lets check: 20^2 + 9^2 = 400 + 81 = 481
16^2 + 15^2 = 256 + 225 = 481. Looks okay.
Lets try another number, say 685.
1--685 = 5 x 137.
2--5 = 2^2 + 1^2 and 137 = 11^2 + 4^2.
.........a........b.............................c........d
3--Using the identity and the designations shown:
For (a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2,
..........................3 x 137 = (2x11 + 1x4)^2 + (2x4 - 1x11)^2
......................................= (22 + 4)^2 + ( 8 - 11)^2
......................................= (26)^2 + (3)^2
Similarly, for (a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2,
.......................................3 x 137 = (2x11 - 1x4)^2 + (2x4 + 1x11)^2
...................................................= ( 22 - 4)^2 + ( 8 + 11)^2
...................................................= (18)^2 + (19)^2
Lets check: 26^2 + 3^2 = 676 + 9 = 685
18^2 + 19^2 = 324 + 361 = 685.
* x^2 + y^2 + z^2 = N^2 has many solutions derived from (a^2 + b^2 - c^2 - d^2) + (2ac + 2bd)^2 + (2ad - 2bc)^2 = (a^2 + b^2 + c^2 + d^2) as long as N is not a power of 2 or 5 times a power of 2.
* No number, prime or composite, of the form 4x - 1 can be expressed as the sum of two squares.
* An odd number of the form N = 4k + 3 cannot be written as the sum of two perfect squares.
* The sum of (n + 1) consecutive squares, beginning with the square of n(2n + 1), is equal to the sum of the
squares of the next n consecutive squares.
Examples:
n = 1, 3^2 + 4^2 = 5^2
n = 2, 10^2 + 11^2 + 12^2 = 13^2 + 14^2
n = 3, 21^2 + 22^2 + 23^2 + 24^2 = 25^2 + 26^2 + 27^2.
n = 4, 36^2 + 37^2 + 38^2 + 39^2 + 40^2 = 41^2 + 42^2 + 43^2 + 44^2
It is worth noting that each series starts with every other triangular number, 3, 10, 21, 36, etc., dropping 1, 6, 15, 28, etc. The last term to the left of the equals sign within each series is the product of 4 times the triangular numbers. The right side of each series has one less term than the left side starting with the last term of the left side plus 1.
* Solutions to x^2 + y^2 = z^4, x, y, and z relatively prime integers, derive from x = 4mn(m^2 - n^2), y = +/-(m^4 - 6m^2n^2 + n^4), and z = m^2 + n^2 where m and n are relatively prime, one odd, one even, m greater than n.
* Three integral squares whose sum is a square as in x^2 + y^2 + z^2 = w^2. Valid solutions can be derived from
x = (p^2 + q^2 - r^2, y = 2pr, z = 2qr, and w = p^2 + q^2 + r^2 where p, q, and r need not necessarily be
rational.
Example: For p = 3, q = 2, r = 1, x = 12, y = 6, z = 4, and w = 14, making 12^2 + 6^2 + 4^2 = 14^2.
* For only 4 pairs of (m,n) does 1^2 + 2^2 + 3^2 + .......+ m^2 = 1 + 2 + 3 + ....+ n; (1,1), (5,10), (6,13), (85,645).
* Squares can be in arithmetic progression - X^2, Y^2 = X^2 + d, Z^2 = Y^2 + d. X = m^2 - n^2 - 2mn, Y = m^2 + n^2, and Z = m^2 - n^2 + 2mn.
Example: m = 3, n = 1, X = 9 - 1 - 6 = 2, Y = 9 + 1 = 10, and Z = 9 - 1 + 6 = 14 and 2^2, 10^2, and 14^2 = 4 - 100 - 196 are in arithmetic progression. Four or more squares cannot be however.
* The sum of the squares of a set of numbers in arithmetic progression equaling a square, x^2 + (x + d)^2 + (x + 2d)^2 + (x + 3d)^2 +.............(x + nd)^2 = N^2.
Example: 2^2 + 5^2 + 8^2 + 11^2 + ...........26^2 = 2304 = 48^2.
* Consecutive squares whose sum is a square.
Examples:
1^2 + 2^2 + 3^2 + .......24^2 = 4900 = 70^2
38^2 + 39^2 + 40^2 + ........48^2 = 20,449 = 143^2
There are many others.
* Would you believe that there are palindomic squares? A few of the smaller ones can be found rather easily.
121, 484, 676, 10201, 12321, 14641, 40804, 69696, 94249, 698896, 1002001 and so on.
In the course of verifying these you will uncover the more amazing fact that some their square roots are also palindromes.
11, 22, 101, 111, 121, 202, 212 and so on.
There are many others.
Some others might be discovered rather easily after inspecting those shown.
101^2 = 10201, 1001^2 = 1002001, 10001^1 = 100020001
202^1 = 40804, 2002^2 = 4008004, 20002^2 - 400080004
Expand the list yourself and see how many more palindromic squares also have palindromic square roots.
* For any three rational numbers, a, b, and c, 1/(a-b)^2 + 1/(b-c)^2 + 1/(c-a)^2 = (m/n)^2, the square of a rational number.
Example: a = 1/3, b = 1/7, c = 1/11
...............a - b = 4/21, b - c = 4/77, c - a = -8/33
..............1/(a-b)^2 = 441/16, 1/(b-c)^2 = 5929/16, 1/(c-a)^2 = 1089/64
..............441/16 + 5929/16 + 1089/64 = 26,569/64 = (163/8)^2.
* The product of a 3 or more digit number and its reverse (as long as they are not equal) is itself a square when both numbers are themselves squares.
Example: 169x961 = 162,409 = 403^2
1089x9801 = 10,673,289 = 3267^2.
* The sum of the reciprocals of the squares yields the unique result of 1 + 1/4 + 1/9 + 1/16 + 25 + .....= Pi^2/6.
* The product of four consecutive numbers of an arithmetic progression of integers plus the 4th power of the common difference is always a perfect square.
Example: From 3, 6, 9, 12, we derive (3x6x9x12) + 3^4 = 1944 + 81 = 2025 = 45^2.
* The successive sums of the squares of the digits in a number will ultimately result in a 1 or a 4 followed by a repeating series of numbers.
Example: Given 8429. 8^2 + 4^2 + 2^2 + 9^2 = 165. 1^2 + 6^2 + 5^2 = 62. 6^2 + 2^2 = 40. 4^2 + 0^2 = 16. 1^2 + 6^2 = 37. 3^2 + 7^2 = 58. 5^2 + 8^2 = 89. 8^2 + 9^2 = 145. 1^2 + 4^2 + 5^2 = 42. 4^2 + 2^2 = 20. 2^2 + 0^2 = 4. 4^2 = 16 > 37 > 58 > 89 > 145 > 42 > 20 > 4 ad infinatum.
* 1^2 = 1 = 0 + 1
2^2 = 4 = 1^2 + 3
3^2 = 9 = 2^2 + 5
4^2 = 16 = 3^2 + 7
5^2 = 25 = 4^2 + 9
6^2 = 36 = 5^2 + 11
* x^2 - y^3 = 1 has but one solution in x = 3 and y = 2.
* Given a, b, and c are rational numbers, [1/(a-b)]^2 + [1/b-c)]^2 + [1/(c-a)]^2 = A^2/B^2.
CUBES
* 1^3 = 0 + 1
2^3 = 3 + 5
3^3 = 7 + 9 + 11
4^3 = 13 + 15 + 17 + 19
* The sum of the first n cubes, i.e., 1^3 + 2^3 + 3^3 +........+ n^3 = (1+2+3+.....+n)^2 = [n(n + 1)/2]^2.
* Three digit numbers that are the sum of the cubes of their digits: 153, 370, 371, 407.
* The smallest number that is the sum of 2 cubes in two different ways. 1729 = 1^3 + 12^3 = 10^3 + 9^3.
The smallest number that is the sum of 2 4th powers in two different ways. 635,318,657 = 59^4 + 158^4 = 133^4 + 134^4.
* Numbers that can be expressed as the sum of two cubes in two different ways are referred to as Hardy-Ramanujan numbers:
1729 = 12^3 + 1^3 = 10^3 + 9^3
4104 = 16^3 + 2^3 = 15^3 + 9^3
13,832 =
20,683 =
32,832 =
39,312 =
* Two cubes with integral sides have their combined volume equal to the combined length of their edges. What
are the dimensions of the cubes? x = 2 and y = 4. Ref. #35 page 87.
* n^3 is always the sum of n odd integers beginning with (n^2 - n + 1) and ending with (n^2 + n - 1).
Example: For n = 6, (n^2 - n + 1) = 31 and 31 + 33 + 35 + 37 + 39 + 41 = 216 = 6^3.
* The sum of a series of three cubes can equal a cube.
Examples:
3^3 + 4^3 + 5^3 = 6^3
11^3 + 12^3 + 13^3 + 14^3 = 20^3
1134^3 + 1135^3 + .........2133^3 = 16,830^3
* Every cube is either a multiple of 9 or right next to one.
SQUBES
There are infiniely many squbes derived from n^6 = n^3xn^3 = (n^3)^2 = n^2xn^2xn^2 = (n^2)^3.
Examples:
1^6 = 1 = 1^2 = 2^3, 2^6 = 64 = 8^2 = 4^3, 3^6 = 729 = 27^2 = 9^3, 4^6 = 4096 = 64^2 = 16^3, 5^6 = 15,625 = 125^2 = 25^3, 6^6 = 46,656 = 216^2 = 36^3, 7^6 = 117,649 = 343^2 = 49^3, etc.
PYTHAGOREAN THEOREM
* Pythagoras is credited with discovering one of the most famous theorems in mathematics. The Pythagorean Theorem states that in any right-angled triangle, the sum of the squares of the two legs adjacent to the right angle is always equal to the square of the longer leg, the hypotenuse. This is expressed by x^2 + y^2 = z^2. Many proofs of this exist, one of which follows. The relationship holds true whether the numbers are integers, rational (fractions) or irrational.
..........I\
..........I \
..........I \
..........I \
......y..I \ z
..........I \
..........I \
..........I________\
x
* One proof: Draw yourself a square ABCD, say 1 1/2 inches square. On each side, locate a point 1 inch from each corner moving clockwise around the square. Label the points E, F, G, and H, E on AB, F on BC, G on CD, and H on DA. Connect points e and F, F and G, G and H, and H and E. You now have a slaller square inside the original square surrounded by four congruent triangles. Label the hypotenuse of each triangle "c" , the short leg of each triangle "b", and the long side of each triangle "a." The area of the inner square is Ai = c^2. The area of the four surrounding triangles is At = 4ab/2 = 2ab. The sum of these areas is the area of the original outer square Ao = 2ab + c^2. The area of the original outer square can also be expressed by Ao = (a + b)^2 = a^2 + 2ab + b^2 which we showed is also equal to 2ab + c^2. Therefore, a^2 + 2ab + b^2 = 2ab + c^2 from which we get a^2 + b^2 = c^2.
* Another proof: In 1876, while serving as a member of the House of Representatives, "future" president James A. Garfield discovered an interesting proof of this theorem. The New England Journal of education published the proof.
The proof involves the use of two methods to calculate the area of a trapezoid. Method one makes use of the formula Area = 1/2(sum of the bases)(altitude) while method two divides the trapezoid into three right triangles and sums the areas of these three triangles.
Construct yourself a trapezoid ABCD with AB parallel to DC, angles C and B right angles. Let AB have length a and DC have length b, b being longer than a. Locate point E on BC such that CE has length a and BE has length b. Draw lines DE and AE and let their lengths be c.
To be more specific, I suggest that you construct the following picture: Draw a horizontal line AB 4 units long, with A at the left end and B at the right end. From point B, draw a vertical line 10 units long, perpendicular to AB, downward, to point C. From point C, draw a horizontal line 6 units long to the left, perpendicular to line BC, to point D. Draw line AD. This is the trapezoid we are talking about. Locate point E on line BC, 4 units up from point C. Then BE = 6 units. Let AB = CE = a and CD = BE = b. Let AE = DE = c.
Notice that angle AED is a right angle also. Now calculate the area of the trapezoid using the two methods identified above.
Area Method 1 = Area Method 2
1/2(a+b)(a+b) = 1/2(ab) + 1/2(ab) + 1/2(cc)
(a+b)(a+b) = (ab) + (ab) + c^2
a^2 + 2ab + b^2 = 2ab + c^2
QED a^2 + b^2 = c^2
Many other proofs may be found at "Pythagorean Theorem Proofs" at the WEB site
http://www.cut-the-knot.com/pythagoras/index.html.
* The altitude to the hypotenuse of a right triangle creates two similar triangles, each similar to the original right triangle and to each other.
* The altitude to the hypotenuse of a right triangle is the geometric mean between the segments of the hypotenuse created by the point where the altitude intersects the hypotenuse or h^2 = xy.
* With an altitude drawn to the hypotenuse of a right triangle, each leg is the geometric mean between the hypotenuse and the segment of the hypotenuse in contact with the leg.
* The altitude to the hypotenuse of a right triangle is derived from h = xy/z
PYTHAGOREAN TRIPLES
* When all three sides of a Pythagorean triangles are integers, they are referred to as Pythagorean Triples. Pythagorean Triples that have no common factor, or a greatest common divisor of 1, are called primitive. Those with a common factor other than 1 are called non-primitive triples.
* All primitive Pythagorean Triples of the form x^2 + y^2 = z^2 derive from x = m^2 - n^2, y = 2mn, and
z = m^2 + n^2 where m and n are arbitrary positive integers of opposite parity (one odd one even), m and n have no common factor, and m is greater than n. (For x, y, & z to be a primitive solution, m and n cannot have common factors and cannot both be even or odd. Violation of these two limitations will produce non-primitive Pythagorean Triples.)
* Triples can also be derived from x = 2n + 1, y = 2n^2 + 2n, and z = 2n^2 + 2n + 1 where n is any integer. The formulas create only triangles where the hypotenuse exceeds the larger leg by one.
* Another set of expressions that produce triples is x = n^2, y = (n^2 - 1)^2/2, and z = (n^2 + 1)^2/2.
* For any positive integer m, 2m, m^2 - 1, and m^2 + 1 are Pythagorean Triples.
* Primitive Pythagorean Triples always have one leg odd, one leg even, and the hypotenuse odd (x and y cannot both be odd or even).
* In every primitive Pythagorean triple, x, y, and z, either x or y is divisible by 3, either x or y is divisible by 4, and either x, y, or z is divisible by 5. The product of the two legs is always divisible by 12, and the product of all three sides is always divisible by 60. The area is always divisible by 6.
* Triples with all sides even are non-primitive but can be reduced to a primitive triple by dividing through by a constant factor.
* Triples with all sides odd are impossible.
* The sum and difference of the hypotenuse and the even side of a primitive Pythagorean triple are squares.
* Only one side of a Pythagorean triangle can be a square.
* If m and n have no factor in common but mn is a square, then both m and n are squares.
* All primitive hypotenuses are primes of the form 4n + 1.
* A number may be a primitive hypotenuse only if all of its prime factors are of the form 4n + 1. No primitive hypotenuse has a prime factor of the form 4n - 1.
* Consecutive values of m and n will produce triples with one leg and the hypotenuse consecutive. An m and n of 2 and 1 produce the 3-4-5 triple. 3 and 2 produce the 5-12-13 triple. 4 and 3 produce the 7-24-25 triple. Make a table of more and see if you can discover the pattern of the results.
* Using the x and y values from the consecutive m and n triples produces triples where the hypotenuse is a perfect square. An m and n of 4 and 3 produces the 7-24-25 triple (25 = 5^2). An m and n of 12 and 5 produces the 119-120-169 triple (169 = 13^2).
* Using the y and z values from the consecutive m and n triples produce triples with the smallest leg a perfect square. An m and n of 5 and 4 produces the 9-40-41 triple. An m and n of 13 and 12 produces the 25-312-313 triple.
* Consecutive legs can be derived from the m and n values derived from the general expression ni = 2(n(p-1) + m(p-2) and mi = 2m(p-1) + n(p-1). The subscript p indicates the "previous" value. The values m(p-1) and n(p-1) are the previous m and n values while m(p-2) and n(p-2) are the values previous to m(p-1) and n(p-1). Consider the starting point of the 3-4-5 triple which derives from m and n values of 2 and 1. The next triple that will produce a triple with both legs consecutive comes from the values of n = 2(1) + 0 = 2 and m = 2(2) + 1 = 5 which produces the 21-20-29 triple. The next will derive from n = 2(2) + 1 = 5 and m = 2(5) + 2 = 12 producing the 119-120-169 triple. See how many you can create and see if you can determine another interesting relationship between the results.
* Other unique Pythagorean Triples can also be derived from m and n values based on the triangular numbers T = n(n+1)/2, i.e., 1, 3, 6, 10, 15, 21, etc. Using consecutive Triangular numbers for m and n, the triples that result have the smallest leg a perfect cube.
* The product of the two legs of a right triangle is equal to the product of the hypotenuse and the altitude to the hypotenuse.
* The area of a Pythagorean triangle is never a square number or twice a square.
There is much more to be observed and learned about Pythagorean triangles which would take much too much room to present here. For instance, it is possible to immediately create a Pythagorean triple given only one integer. There are many problems in Pythagorean triangles that have engaged the attention of mathematicians for centuries, such as, "If A is a given integer, of how many primitive Pythagrorean triangles may it be a leg?" or "Find three or more Pythagorean triangles having equal areas." Then there are problems involving Pythagorean triangles whose areas have a given ratio; whose area equals its perimeter; whose perimeter is a square; the sum of whose area and perimeter is a cube; whose area equals the hypotenuse; whose perimeter is a given number; and so on, ad infinatum.
It was during the course of the creation of such problems that mathematicians finally met their match when they tried to find those Pythagorean triangles whose areas were squares. Pierre Fermat finally proved it was impossible.
Another interesting problem that might challenge you is to find the smallest solution, in integers, of the inverse Pythagorean Triple relationship
1/x^2 + 1/y^2 = 1/z^2.
Finding Pythagorean triples with specific integers.
* Given any integer one can find another such that the sum of their squares is an integral square. Given the number A is odd - Resolve A into any 2 of its factors, equating the larger to m + n and the smaller to M - n. Solve for m and n and substitute into the formulas of 1) above. Given the number A is even - equate it to mn and let m and n be any two integers which will result in the product being eqaul to A.
* Pierre Fermat discovered that Pythagorean Triples can be derived by expressing the odd squares as the sum of consecutive integers. Using any odd square, express it as x + (x + 1) = N = n^2 and n^2 + x^2 = (x + 1)^2.
Another way of looking at it is to say a^2 = (b + c), b and c being consecutive integers. Since (b - c) = 1, we
can write a^2 = (b + c)(b - c) = b^2 - c^2 from which a^2 + b^2 = c^2. For example, 25 = 12 + 13 = 5^2 and
5^2 + 12^2 = 13^2 or 121 = 60 + 61 = 11^2 and 11^2 + 60^2 = 61^2.
* Others can be derived by taking any consecutive odd or even numbers and adding their reciprocals as with x
and y, we get 1/x + 1/y = (x + y)/xy. It follows that (x + y)^2 + (xy)^2 = N^2. Example, from 5 and 7 we get
1/5 + 1/7 = 12/35 and 12^2 + 35^2 = 37^2 or from 6 and 8 we get 1/6 + 1/8 = 7/24 and 7^2 + 24^2 = 25^2.
* Of how many primitive Pythagorean triangles may N be a leg?
If a number N is made up from n primes and their powers, it is the leg of 2^(n - 1) primitive Pythagreaon
triangles. If N = 2700, its 3 factors, 2^2(3^3)5^2, give rise to 2^(3-1) = 4 primitive triangles. (If N is even but not
divisible by 4, there are no solutions, as there cannot be for any other number of the form 4x + 2.)
* Of how many primitive and non-primitive Pythagorean triangles may N be a leg?
Given N = 2^(ao)(p1^(a1))(p2^(a2))....pn^(an), the number of triangles N can be a leg of is given by
Legs = [(2(ao) - 1)(2(a1) + 1)(2(a2) + 1)....(2(an) + 1) - 1]/2.
Example: For N = 2700, where N = 2^2(3^3)5^2, the number of possible triangles with N = 2700 is
L = [(2(2) - 1)(2(3) + 1)(2(2) + 1) -1]/2 = 52.
* If a number N has a total of n distinct prime divisors, each being of the form 4x + 1, then N can be the
hypotenuse of 2^(n - 1) primitive Pythagorean triangles.
..............................a1 a2 an b1 b2 br
* If the number N = p(1) x p(2) x..... p(n) x q(1) x q(2) x .....q(r) where p's are primes of the form 4x - 1 and q's
are of the form 4x + 1, N cannot be the hypotenuse of a single primitive Pythagorean triangle but may be the
hypotenuse of H = [(2b1 + 1)(2b2 + 1).....(2br + 1) - 1]/2 non-primitive ones. The a's and b's are the powers to
which the p's and q's are raised.
* How many primitive Pythagorean triples, a, b, c, are there where (x + y) is itself a square, d^2?
Calculate d from d = p^2 - q^2/2, m from m = p^2 - pq + q^2/2, and n from n = d + pq - p^2 + q^2/2 where p is odd and q is even.
Example: For p = 5 and q = 2, d = 23, m = 17, n = 10, making x = m^2 - n^2 = 189, y = 2mn = 340, and z = m^2 + n^2 = 389 from which 189 + 340 = 529 = 23^2.
* Given any integer one can find another integer such that the sum of their squares is an integral square.
Given the number A is odd - Resolve A into any 2 of its factors, equating the larger to m + n and the smaller to m - n. Solve for m and n and substitute into the formulas of 1) above.
Given the number A is even - equate it to 2mn and let m and n be any two integers which will result in the product 2mn being equal to A.
* Given an integer, it is possible to derive a Pythagorean triple with that integer as one of the sides. For odd integers, given the integer X; resolve X into any two factors; equate the larger factor to m + n and the smaller to m - n; solve for m and n; define the triple using the above expressions. For even integers, given the integer X; equate X to 2mn; let m and n be any two numbers that yield the product 2mn; to derive primitive solutions, X must be relatively prime.
* Some Pythagorean Triples can be derived using the odd squares. Using any odd square, express it as the sum of two consecutive integers, as in x + (x + 1) = N = n^2 and n^2 + x^2 = (x + 1)^2. For example, 25 = 12 + 13 = 5^2 and 5^2 + 12^2 = 13^2 or 121 = 60 + 61 = 11^2 and 11^2 + 60^2 = 61^2.
* Other triples can be derived by taking any consecutive odd or even numbers and adding their reciprocals as with x and y, we get 1/x + 1/y = (x + y)/xy. It follows that (x + y)^2 + (xy)^2 = N^2. Example, from 5 and 7 we get 1/5 + 1/7 = 12/35 and 12^2 + 35^2 = 37^2 or from 6 and 8 we get 1/6 + 1/8 = 7/24 and 7^2 + 24^2 = 25^2.
* Solutions to x^2 + y^2 = z^4, x, y, and z relatively prime integers, derive from x = 4mn(m^2 - n^2), y = +/-(m^4 - 6m^2n^2 + n^4), and z = m^2 + n^2 where m and n are relatively prime, one odd, one even, m greater than n.
* Rational solutions of x^2 + y^2 = 1 derive from x = 2r/(1+r^2) and y = (1 - r^2)/(1+r^2) where r = m/n, m and n relatively prime integers, m greater than n.
* In spite of the multitude of formulas and methods for deriving Pythagorean triples, and the number of triples that result, any pair of such triangles are surprisingly related to one another. If we label our triangle legs and hypotenuses by X, Y, Z and x, y, z, respectively, it will surprise you to learn that (Z + z)^2 - (X + x)^2 - (Y + y)^2 = W^2, another square. Also noteworthy are the relationships between all the legs of Zz - Xx - Yy = 2A^2, Zz + Xx + Yy = 2B^2, Zz - Xx + Yy = 2C^2, Zz + Xx - Yy = 2D^2, Zz - Xy - xY = E^2, Zz + Xy + xY = F^2, Zz + Xy - xY = G^2, and Zz - Xy + xY = H^2.
* Pythagorean Triple triangles with integral sides and equal areas: 40-42-58, 24-70-74, and 15-112-113.
* If you enjoy working/playing with Pythagorean Triples, you might also like to experiment with 3 dimensional
Pythagorean Triples.
1--How many rectangular solids can you define where all the edges and the surface diagonals are all integers?
Picture a solid rectangular block with dimensions of x, y, and z and face diagonals of X, Y, and Z. Thus (y^2
+ z^2) = X^2, (x^2 + z^2) = Y^2, and (x^2 + y^2) = Z^2.
Using any primitive right triangle with sides a, b, and c as a reference where a = m^2 - n^2, b = 2mn, and c =
m^2 + n^2, m and n being any two integers, one being odd and the other being even.
With a, b, and c identified, the block sides are then defined by x = a(4b^2 - c^2), y = b(4a^2 - c^2), and z =
4abc. Similarly, the diagonals are defined by X = b(4a^2 + c^2), Y = a(4b^2 + c^2), and Z = c^3.
These formulas do not give all the possible solutions. Some not derived from the formulas are:
x...85.....187.....195.....231.....275.....855.....1105.....1155.....1155
y..132...1020....748.....792.....252.....2640....9360.....1100.....6300
z..720...1584...6336....160.....240......832....35904....1008.....6688
2--A. H. Beiler, in his absolutely wonderful book Recreations in the Theory of Numbers, Dover Publications,
Inc., 1964, states that it is possible to have three integral squares, X, Y, Z, such that their sums, in
pairs, are also squares, i.e., X^2 + Y^2 = a^2, X^2 + Z^2 = b^2, and Y^2 + Z^2 = c^2. This literally reflects
the same problem in that the expressions represent a box of dimensions X, Y, and Z with surface diagonals
of a, b, and c. Using the standard Pythagorean Triples expressions again, he derived some general formulas
giving solutions (but not all) as follows:
X = 2mn(3m^2 - n^2)(3n^2 - m^2) a = 2mn(5m^4 - 6m^2n^2 + 5n^4)
Y = 8mn(m^4 - n^4) b = (m^2 + n^2)^3
Z = (m^2 - n^2)(m^2 + n^2 + 4mn)(m^2 + n^2 - 4mn) c = (m^2 - n^2)(m^4 + 18m^2 n^2 + n^4)
where m and n are arbitrary numbers, m being greater than n. Thus, a box of dimensions 44x117x240 has
surface diagonals of 125, 244, and 267, derived from m = 2 and n = 1. Many other such triplets have been
found through the use of computer programs.
3--How many rectangular solids can you define where all the edges and the internal diagonal are all integers?
Here, we have to satisfy the expression X^2 + Y^2 + Z^2 = d^2 where d = the internal diagonal. These can
be derived from the expressions:
........X = p^2 + q^2 - r^2
.......Y = 2pr
.......Z = 2qr
.......d = p^2 + q^2 + r^2
One immediately obvious solution is 1^2 + 2^2 + 2^2 = 3^2. Some others are
X.....Y.....Z.....d
2.....3.....6.....7
1.....4.....8.....9
3....16...24...29 (Note - p, q, and r do not necessarily have to be rational.)
4--As for the ultimate 3-dimensional Pythagorean Triple box, how many rectangular solids can you define
where all the edges, all the surface diagonals, and the internal diagonal are all integers?
To find such a box the expressions from above give rise to Y^2 + b^2 = d^2 or d^2 = [8mn(m^4 - n^4)]^2 +
[(m^2 + n^2)^3]^2. A computer program can no doubt be created that will search for the smallest integer
value of d, given one exists. I have found no solutions to date and do not, in fact, know if any exist.
* Can three Pythagorean triangles be arranged such that their legs form the sides of a rectangle and their hypotenuses form another triangle internal to the rectangle? Try fitting together the Pythagorean Triples of 9-12-15, 12-16-20, and 7-24-25 into such an arrangement. How many others can you identify?
* Solutions to x^2 + y^2 = z^4 may be derived from x = 4mn(m^2 - n^2), y = +/-(m^4 - 6m^2n^2 + n^4), and z = (m^2 + n^2), m > n, m and n are relatively prime, with one odd and one even.
n^2+2n+1) - n^2 \:=\:2n+1,\:\text{ an odd number.}\)