Divisibility!

Number Theory Level 5

$P=(a^{3} -b^{3})(b^{3} -c^{3})(c^{3} -d^{3})(d^{3} -a^{3})(c^{3} -a^{3})(d^{3} -b^{3})$

If $a$ , $b$ , $c$ , and $d$ are four positive integers then $P$ is always divisible by a square integer.

Find the maximum value of this integer.

1 solution

Kushal Bose
Jun 21, 2016

Let $\large x\equiv 0,1,2,3,4,5,6,7,8\pmod 9$

So $\large x^{3} \equiv 0,1,8 \pmod 9$

In a set of four integers $a,b,c,d$ pairwise differences of their cubes there exists a single pair whose difference will be divisible by 9

Therefore, the highest perferct square factor is 9.

Your argument does not show that 9 is the highest possible number. In fact, 144 is the answer.

Jon Haussmann - 4 years, 11 months ago

I have written this solution before you posted your solution.

Kushal Bose - 4 years, 11 months ago