a=b=p is a solution, I’m not sure if this problem gives enough information

Hopefully my solution is correct. We know :- $ab|a^3+b^3 \implies ab|(a+b)(a^2+b^2)$ $ab|4(a^2+b^2)+p(a+b) \implies ab|p(a+b)^2$ by multiplying RHS with $(a+b)$. Which $\implies ab|p(a^2+b^2)$ Let $d$ be HCF of $a,b$. Therefore let $a=dx, b=dy$ where $HCF(x,y)=1$. Thus plugging the values of $a,b$ we arrive at $xy|p(x^2+y^2) \implies x|p(x^2+y^2) \implies x|py^2$ Let $q$ be a prime which divides $x$. Then $q|x|py^2 \implies q|p \implies q=p$ because q can’t divide $y^2$ as it would contradict the HCF=1 fact. But similarly this implies that $p|y$ which again contradicts the HCF=1 fact. Thus atleast one of $x$ or $y$ equals $1$. WLOG, let $y=1$ $\implies x|p \implies x=1$ OR $x=p$ $\implies a=bp, b=ap, a=b $ are the only solutions. @below thank you for the correction, I’ve fixed it.

Gumnaami_1945 wrote: Hopefully my solution is correct. We know :- $ab|a^3+b^3 \implies ab|(a+b)(a^2+b^2)$ $ab|4(a^2+b^2)+p(a+b) \implies ab|(a+b)^2$ by multiplying RHS with $(a+b)$. Which $\implies ab|a^2+b^2$ Let $d$ be HCF of $a,b$. Therefore let $a=dx, b=dy$ where $HCF(x,y)=1$. Thus plugging the values of $a,b$ we arrive at $xy|x^2+y^2 \implies x|x^2+y^2 \implies x|y^2$ Let $q$ be a prime which divides $x$. Then $q|x|y^2 \implies q|y$ which contradicts the fact that $HCF(x,y)=1$. Thus no prime can divide $x$ or $y \implies x=y=1 \implies a=b$ which is necessary but not sufficient, because now we need to show $ab|p(a+b) \implies a|2p \implies a=b=1,2,p,2p$ satisfies. We don't always have $a=b$. Notice that $a=5,b=25$ works for $p=5$. We only have $ab|p(a+b)^2$ from $ab|4(a^2+b^2)+p(a+b)$.