Prove any set S of three integers contains a pair $x\neq y$ such that $x^3y-xy^3$ is divisible by 10.

Question

Prove any set S of three integers contains a pair $x\neq y$ such that $x^3y-xy^3$ is divisible by 10.

My attempt was : By the division algorithm, every integer $n$ can be written as $n = 10q + r,$ where $0\leq r \leq 9$. Since there are three integers in $S$ but only 10 possible values for the remainder r. By the pigeonhole principle, the set S contains x & y that leave the same remainder on division by 10, that is there exist s with $0\leq s \leq 9$ such that $x = 10q_1 +s$ and $y = 10q_2+s.$ From here plug x and y into $x^3y-xy^3$ and show there is a 10 I can factor out.

Am I going about this problem the right way?

Answer 1

The problem with your approach has been pointed out in the comments by lulu.

Note that $x^3y-xy^3=xy(x^2-y^2)=xy(x-y)(x+y)$.

• Show that at least one of the four factors must be even no matter what $x$ and $y$ are.

If $x$ or $y$ is a multiple of $5$, or if $x\equiv y\pmod5$, or if $x+y\equiv0\pmod 5$, then $x^3y-xy^3$ is a multiple of $10$. What combinations of $x$ and $y$ does that cover if neither is a multiple of $5$? Look at $x$ and $y$ modulo $5$:

$$\begin{array}{c|cc} x\backslash y&1&2&3&4\\\hline 1&\checkmark&&&\checkmark\\ 2&&\checkmark&\checkmark\\ 3&&\checkmark&\checkmark\\ 4&\checkmark&&&\checkmark \end{array}$$

• Show that any set of three integers must contain a multiple of $5$ or two integers in one of that form one of the covered combinations.