2021 USA TST #1

Determine all integers $s \ge 4$ for which there exist positive integers $a$, $b$, $c$, $d$ such that $s = a+b+c+d$ and $s$ divides $abc+abd+acd+bcd$.

The answer is $\boxed{\text{all composites }s}$.

For sufficiency, if $s$ is composite, we can write $s = xy$ and take $(a, b, c, d) = (1, x - 1, y - 1, (x - 1)(y - 1))$, which works because then\[abc + abd + acd + bcd = xy(x - 1)(y - 1).\] 

Now for necessity, if $s$ is prime and $s\mid abc+abd+acd+bcd$, then:

$$0\equiv abc+abd+acd+bcd$$

$$\equiv a(bc+bd+cd)+bcd$$

$$\equiv bcd-(b+c+d)(bc+bd+cd)$$

$$\equiv -(b^2c+bc^2+c^2d+cd^2+d^2b+db^2+2bcd)$$

$$\equiv-(b+c)(c+d)(d+b)\pmod s.$$

Either $s\mid b+c$, $s\mid c+d$, or $s\mid d+b$, but since $s$ is greater than all of these, we have a contradiction. 

So, $s$ must be composite, as desired. $\square$


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