On $a^4 + b^4 = c^4 + d^4 = e^5$. 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.
 A: 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.
A: I used Beal's conjecture to show that $ac+bd$ is composite.
We have $a^4+b^4=e^5$, Using Beal's conjecture we say that $a$ and $b$ will have common prime factor.
Let the common prime factor be $f$.
$a=a_1f$ and $b=b_1f$ for some $a_1,b_1\in \mathbb{Z^+}$
In $ac+bd=a_1fc+b_1fd=f(a_1c+b_1d)$
Therefore $ac+bd$ is composite.
Above was a proof using a conjecture. I tried of proving it without the use of any conjecture but I failed to prove. 
I was able to prove that $ac+bd$ is composite when $e$ is even.
$e$ is even, then $e^5$ will be even, then $a^4+b^4$ will be even, then $a+b$ will be even.
Therefore either both $a$ and $b$ are odd or both are even.
Similarly we will have either $c$ and $d$ odd or both even.
Now with any combination we will get $ac+bd$ as even.
Therefore $ac+bd$ is composite.
When doing the same analysis when taking $e$ as odd
Then we will get that either $a$ or $b$ is even and either $c$ or $d$ is even.
When $a$ and $c$ are odd and $b$ and $d$ are even then this may be possible that $ac+bd$ is prime or this may be composite.(yet I don't have any proof of this) This is the reason why I used the conjecture. 
