Prove that the difference of x^3+y and y^3+x is a always a multiple of 6

[willr12]

I recently came across a problem that said 'prove that the difference of x^3+y and y^3+x is a always a multiple of 6, given that x and y are integers'. I tried it, but I'm not sure if this is a valid proof.

this is with the first triangular number being 0 and Tsubn being the nth triangular number.

 

[mfb]

If you can use or prove the formula for the triangular numbers, sure.

 

[Mark44]

Or you can do this:
x³ + y - (y³ + x) = x³ - x - (y³ - y)
= x(x² - 1) - y(y² - 1)
= x(x - 1)(x + 1) - y(y - 1)(y+ 1)

In a more suggestive form, the last expression is
(x - 1)(x)(x + 1) - (y - 1)(y)(y + 1), with each group being the product of three successive integers.

It's not too hard to show that the product of three integers in succession is divisible by 6 (can do this by induction), so the difference of two such products is also divisible by 6.

 

[mfb]

(can do this by induction)

Or splitting it into 5 to 6 cases.
That is the more elementary approach if the triangle formula is not available.

 

[alan2]

No need to consider different cases . Given 3 consecutive integers, exactly one is a multiple of 3 and at least one is even. Therefore the product is divisible by both 2 and 3.