ab(a^2-b^2) conundrum
- Thread starter gortwell
- Start date
Here is what I read (got out of) in the two previous posts:
gortwell, you used a double negative, so it means out of all of your examples you looked at
it is always true.
Then, srmichael, your example is not a counterexample to gortwell's because it is true.
gortwell either wants to know if it's always true (proof?) or is there a counterexample
to show it's not always true.
Hi,
sorry for the ambiguity
I didn't not mean to do it
I was trying to find out if \(\displaystyle ab(a^2-b^2)\) is always of the form 6k.
Using modular arithmetic, the above proves to be true
sorry for the ambiguity
I didn't not mean to do it I was trying to find out if \(\displaystyle ab(a^2-b^2)\) is always of the form 6k.
Using modular arithmetic, the above proves to be true
| a Mod 6 | b Mod 6 | ab Mod 6 | a+b Mod 6 | a-b Mod 6 | \(\displaystyle ab(a^2-b^2)\) Mod 6 |
| 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 5 | 0 |
| 0 | 2 | 0 | 2 | 4 | 0 |
| 0 | 3 | 0 | 3 | 3 | 0 |
| 0 | 4 | 0 | 4 | 2 | 0 |
| 0 | 5 | 0 | 5 | 1 | 0 |
| 1 | 0 | 0 | 1 | 1 | 0 |
| 1 | 1 | 1 | 2 | 0 | 0 |
| 1 | 2 | 2 | 3 | 5 | 0 |
| 1 | 3 | 3 | 4 | 4 | 0 |
| 1 | 4 | 4 | 5 | 3 | 0 |
| 1 | 5 | 5 | 0 | 2 | 0 |
| 2 | 0 | 0 | 2 | 2 | 0 |
| 2 | 1 | 2 | 3 | 1 | 0 |
| 2 | 2 | 4 | 4 | 0 | 0 |
| 2 | 3 | 0 | 5 | 5 | 0 |
| 2 | 4 | 2 | 0 | 4 | 0 |
| 2 | 5 | 4 | 1 | 3 | 0 |
| 3 | 0 | 0 | 3 | 3 | 0 |
| 3 | 1 | 3 | 4 | 2 | 0 |
| 3 | 2 | 0 | 5 | 1 | 0 |
| 3 | 3 | 3 | 0 | 0 | 0 |
| 3 | 4 | 0 | 1 | 5 | 0 |
| 3 | 5 | 3 | 2 | 4 | 0 |
| 4 | 0 | 0 | 4 | 4 | 0 |
| 4 | 1 | 4 | 5 | 3 | 0 |
| 4 | 2 | 2 | 0 | 2 | 0 |
| 4 | 3 | 0 | 1 | 1 | 0 |
| 4 | 4 | 4 | 2 | 0 | 0 |
| 4 | 5 | 2 | 3 | 5 | 0 |
| 5 | 0 | 0 | 5 | 5 | 0 |
| 5 | 1 | 5 | 0 | 4 | 0 |
| 5 | 2 | 4 | 1 | 3 | 0 |
| 5 | 3 | 3 | 2 | 2 | 0 |
| 5 | 4 | 2 | 3 | 1 | 0 |
| 5 | 5 | 1 | 4 | 0 | 0 |
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Hi JeffM, thanks for the contribution ;-)
Another way would be this...
\(\displaystyle ab(a^2-b^2)\equiv ab(a+b)(a-b)\)
if a and b are both even then all 3 factors are even \(\displaystyle \therefore ab(a^2-b^2)\) is even
if a and b are both odd then (a+b) and (a-b) are even \(\displaystyle \therefore ab(a^2-b^2)\) is even
if a and b are opposite parity then ab is even \(\displaystyle \therefore ab(a^2-b^2)\) is even
This proves that \(\displaystyle ab(a^2-b^2)\) is even (i.e. a multiple of 2)
All we need now is to prove \(\displaystyle ab(a^2-b^2)\) is also a multiple of 3
If either a or b is divisible by 3 (0 Mod 3) then \(\displaystyle ab(a^2-b^2)\) is divisible by 3
If a and b have the same remainder when divided by 3 then (a-b) = 0 Mod 3 so \(\displaystyle ab(a^2-b^2)\) is divisible by 3
The only remaining possibility is a and b are equal to 1 Mod 3 and 2 Mod 3 (or vice versa)
In this case (a+b)=0 Mod 3
So \(\displaystyle ab(a^2-b^2)\) is divisible by 2 and divisible by 3, therefore it is divisible by 6
Another way would be this...
\(\displaystyle ab(a^2-b^2)\equiv ab(a+b)(a-b)\)
if a and b are both even then all 3 factors are even \(\displaystyle \therefore ab(a^2-b^2)\) is even
if a and b are both odd then (a+b) and (a-b) are even \(\displaystyle \therefore ab(a^2-b^2)\) is even
if a and b are opposite parity then ab is even \(\displaystyle \therefore ab(a^2-b^2)\) is even
This proves that \(\displaystyle ab(a^2-b^2)\) is even (i.e. a multiple of 2)
All we need now is to prove \(\displaystyle ab(a^2-b^2)\) is also a multiple of 3
If either a or b is divisible by 3 (0 Mod 3) then \(\displaystyle ab(a^2-b^2)\) is divisible by 3
If a and b have the same remainder when divided by 3 then (a-b) = 0 Mod 3 so \(\displaystyle ab(a^2-b^2)\) is divisible by 3
The only remaining possibility is a and b are equal to 1 Mod 3 and 2 Mod 3 (or vice versa)
In this case (a+b)=0 Mod 3
So \(\displaystyle ab(a^2-b^2)\) is divisible by 2 and divisible by 3, therefore it is divisible by 6
Hello, gortwell!
We have: .\(\displaystyle N \:=\:ab(a-b)(a+b)\)
We must show that \(\displaystyle N\) is a multiple of \(\displaystyle 2\) and a multiple of \(\displaystyle 3.\)
\(\displaystyle N\) is the product of four integers.
If one factor is even, then \(\displaystyle N\) is even, a multiple of 2.
. . If \(\displaystyle a\) or \(\displaystyle b\) is even, then \(\displaystyle N\) is even.
. . If \(\displaystyle a\) and \(\displaystyle b\) are both odd, then \(\displaystyle (a+b)\) is even.
Hence, \(\displaystyle N\) is a multiple of 2.
If one factor is a multiple of 3, then \(\displaystyle N\) is a multiple of 3.
. . If \(\displaystyle a\) or \(\displaystyle b\) is a multiple of 3, then \(\displaystyle N\) is a multiple of 3.
. . If neither \(\displaystyle a\) nor \(\displaystyle b\) is a multiple of 3, we have four cases to consider.
. . . . \(\displaystyle a \,=\,3m \pm1,\;b \,=\,3n \pm1\)
We have the following scenarios:
\(\displaystyle \begin{array}{|c|c|c|}a & b & (a-b)& (a+b) \\ \hline
3m+1 & 3n+1 &3(m-n) & -- \\
3m+1 & 3n-1 & -- & 3(m-n) \\
3m-1 & 3n+1 & -- & 3(m+n) \\
3m-1 & 3n-1 & 3(m-n) & -- \end{array}\)
In every case, \(\displaystyle N\) is a multiple of 3.
Therefore, \(\displaystyle N\) is a multiple of 6.
We have: .\(\displaystyle N \:=\:ab(a-b)(a+b)\)
We must show that \(\displaystyle N\) is a multiple of \(\displaystyle 2\) and a multiple of \(\displaystyle 3.\)
\(\displaystyle N\) is the product of four integers.
If one factor is even, then \(\displaystyle N\) is even, a multiple of 2.
. . If \(\displaystyle a\) or \(\displaystyle b\) is even, then \(\displaystyle N\) is even.
. . If \(\displaystyle a\) and \(\displaystyle b\) are both odd, then \(\displaystyle (a+b)\) is even.
Hence, \(\displaystyle N\) is a multiple of 2.
If one factor is a multiple of 3, then \(\displaystyle N\) is a multiple of 3.
. . If \(\displaystyle a\) or \(\displaystyle b\) is a multiple of 3, then \(\displaystyle N\) is a multiple of 3.
. . If neither \(\displaystyle a\) nor \(\displaystyle b\) is a multiple of 3, we have four cases to consider.
. . . . \(\displaystyle a \,=\,3m \pm1,\;b \,=\,3n \pm1\)
We have the following scenarios:
\(\displaystyle \begin{array}{|c|c|c|}a & b & (a-b)& (a+b) \\ \hline
3m+1 & 3n+1 &3(m-n) & -- \\
3m+1 & 3n-1 & -- & 3(m-n) \\
3m-1 & 3n+1 & -- & 3(m+n) \\
3m-1 & 3n-1 & 3(m-n) & -- \end{array}\)
In every case, \(\displaystyle N\) is a multiple of 3.
Therefore, \(\displaystyle N\) is a multiple of 6.
\(\displaystyle ab(a^2 - b^2) \ = \ \)
\(\displaystyle ab(a^2 - b^2 \ - 1 + 1) \ = \)
\(\displaystyle ab(a^2 - 1 \ - \ b^2 + 1) \ =\)
\(\displaystyle ab[(a^2 - 1) \ - \ (b^2 - 1)] \ = \ \)
\(\displaystyle ab(a^2 - 1) \ - \ ab(b^2 - 1) \ = \)
\(\displaystyle ab(a - 1)(a + 1) \ - \ ab(b - 1)(b + 1) \ = \)
\(\displaystyle b(a - 1)(a)(a + 1) \ - \ a(b - 1)(b)(b + 1) \)
Each product is the product of at least two consecutive
integers, so each is even. Each product is the product
of three consecutive integers, so each is divisible by 3.
Because each product is divisible by 2 and by 3,
then each product is a multiple of 6.
The original expression is equal to the difference of
two multiples of 6, so the original product is itself a
multiple of 6.
So, \(\displaystyle \ ab(a^2 - b^2) \ = \ 6k, \ \ \ where \ \ k \ \ is \ \ an \ \ integer.\)
\(\displaystyle ab(a^2 - b^2 \ - 1 + 1) \ = \)
\(\displaystyle ab(a^2 - 1 \ - \ b^2 + 1) \ =\)
\(\displaystyle ab[(a^2 - 1) \ - \ (b^2 - 1)] \ = \ \)
\(\displaystyle ab(a^2 - 1) \ - \ ab(b^2 - 1) \ = \)
\(\displaystyle ab(a - 1)(a + 1) \ - \ ab(b - 1)(b + 1) \ = \)
\(\displaystyle b(a - 1)(a)(a + 1) \ - \ a(b - 1)(b)(b + 1) \)
Each product is the product of at least two consecutive
integers, so each is even. Each product is the product
of three consecutive integers, so each is divisible by 3.
Because each product is divisible by 2 and by 3,
then each product is a multiple of 6.
The original expression is equal to the difference of
two multiples of 6, so the original product is itself a
multiple of 6.
So, \(\displaystyle \ ab(a^2 - b^2) \ = \ 6k, \ \ \ where \ \ k \ \ is \ \ an \ \ integer.\)