justin said:
Difference of Cubes? (a^3 - b^3)
also (a^4 - b^4)
what do these equal
They could be set equal to anything basically. For instance:
Find two perfect cubes whose difference is a perfect square.
Letting x = a, y = ma (m > 1) and z = na, we can write a^3(m^3 - 1) = n^2(a^2) or a = n^2/(m^3 - 1).
We need to find compatible integer values of "m" and "n" that produce integer values for "a".
For a given "m", search for values of "n" that result in n^2 being evenly divisible by (m^3 - 1) producing "a".
Alternatively, for m = 2 on up, a = n^2/7 for m = 2, = n^2/26 for m = 3, = n^2/63 for m = 4, etc.
m......n--->...1.....2.....3.....4.....5.....6.....7.....8.....9.....10.....11.....12.....13.....14.....15
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1...............................................................................................................................none
2.........................................................7....................................................28.............from n = 7x1, 2, 3,...i
3...............................................................................................................................from n = 26x1, 2, 3,...i
Values of "n" for higher values of "m" must be derived in the same manner as those given above.
Check: For n = 7 and m = 2, a = 7^2/(8 - 1) = 7 making x = 7, y = 14 and z = 49.
Then, 14^3 - 7^3 = 49^2 or 2744 - 343 = 2401.
y^4 - x^4 = z^3
FInd two numbers, the difference of whose fourth powers is a perfect cube.
Let x = a, y = ma and z = na.
Then, m^4a^4 - a^4 = n^3a^3 or a = n^3/(m^4 - 1).
By inspection, m must be greater than 1 and therefore, m = 1 and n = an even number cannot lead to integer answers for x, y and z.
With m > 1, all answers are rational with no integer answers being possible.
Example:
Letting m = 2 and n = 5 makes a = 125/15.
Then, x = a = 125/17, y = 250/17 and z = 625/17.
Then, (250/15)^4 - (125/15)^4 = (625/15)^3.