m^3n - mn^3 mod 6 = 0

[tmoria]

Hi,

I know that \(\displaystyle m^3n-mn^3\) can be expanded to \(\displaystyle mn(m+n)(m-n)\) but cannot figure why the result for any \(\displaystyle m\) and \(\displaystyle n\ \ =0\ \ mod\ \ 6\)

Any ideas?

 

[stapel]

I know that \(\displaystyle m^3n-mn^3\) can be expanded to \(\displaystyle mn(m+n)(m-n)\) but cannot figure why the result for any \(\displaystyle m\) and \(\displaystyle n\ \ =0\ \ mod\ \ 6\)

Any ideas?

If m = n, then m - n = 0-mod-(anything). If either of m and n equals 0, then mn = 0-mod-(anything).

Suppose m does not equal n, and suppose that neither is zero.

If m = 1 and n = 2, then m + n = 3 and n(m + n) = 2(3) = 6 = 0-mod-6. (If n = 1 and m = 2, the result is obviously the same.)

If m = 1 and n = 3, then m - n = -2 = 4-mod-6 = 2*2-mod-6, so n(m - n) = 3(2*2) = 6*2 = 0*2-mod-6 = 0.

And so forth.

 

[tmoria]

Thanks stapel