Let $a, b, c, d, e$ be distinct positive integers such that $a^4 + b^4 = c^4 + d^4 = e^5$. Show that $ac + bd$ is a composite number.

# [Math] On $a^4 + b^4 = c^4 + d^4 = e^5$.

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# [Math] On $a^4 + b^4 = c^4 + d^4 = e^5$.

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Let $a, b, c, d, e$ be distinct positive integers such that $a^4 + b^4 = c^4 + d^4 = e^5$. Show that $ac + bd$ is a composite number.

## Best Answer

Assume that $p=ac+bd$ is prime. Without loss of generality, we may assume that $a>c,d$ from which follows that $b<c,d$. Also, $a,b,c,d<p$, and so must all be relatively prime with $p$.

Computing modulo $p$, we have $ac \equiv -bd$ which leads to $(ac)^4\equiv(bd)^4$. Exploiting $a^4+b^4=c^4+d^4$ we get $$ (a^4+b^4)c^4=(ac)^4+(bc)^4\equiv(db)^4+(bc)^4=(c^4+d^4)b^4=(a^4+b^4)b^4 $$ which means that either $p$ divides $a^4+b^4=c^4+d^4$, or $b^4\equiv c^4$.

We can exclude the alternative $p|a^4+b^4$ since that would make $p|e$, which in turn would require $e\ge p=ac+bd>a+b$ which leads to $e^5>a^4+b^4$. Note that this is the only use for the equality with $e^5$.

So now we know that $b^4\equiv c^4$, which means that $$ p|c^4-b^4=(c^2+b^2)(c+b)(c-b). $$ Again, $p>c+b$, so we must have $p|c^2+b^2$. However, $c^2+b^2<ac+bd=p$ as $c<a$ and $b<d$, which makes $p|c^2+b^2$ impossible.

Thus, no such $a,b,c,d$ exist for which $p$ is prime.

This was written late at night, so I hope I didn't make any glaring mistakes.