<s>[INST] Find all solutions to $a^3 + b^3 = p^4,$ where $p$ is prime, and $a,b$ are natural Initially, I tried to break the equation down into $(a+b)(a^2-ab+b^2)$, then break it down into two cases:
I: $(a+b) = p, (a^2-ab+b^2) = p^3$, which yields to a contradiction that $p^2 > p^3.$
II: $(a+b) = p^2, (a^2-ab+b^2) = p^2$, which converts to $p^2(p-1)(p+1) = 3ab.$
However, I can't find any way to use this.
 [/INST] You are on the right track when you render $a+b=p^2,a^2-ab+b^2=p^2$.
You should then plug in
$a^2-ab+b^2=\frac14(a^2+2
ab+b^2)+\frac34(a^2-2ab+b^2)$
$=\frac14(a+b)^2+\frac34(a-b)^2.$
Then conclude that with $a+b=p^2,a^2-ab+b^2=p^2$, you must have
$p^2\ge\frac14p^4.$
From this $p\le2$, thereby limiting your solution candidates.
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