Equation $a^5+15ab+b^5=1$ What are the integer solutions of $a^5+15ab+b^5=1$?
The equation is symmetric in $a$ and $b$, so let's assume $a\geq b$. When $a=b$, we have $2a^5+15a^2=1$, which has no solution by the Rational Root Theorem. Else, $a>b$. It is degree-five, so we cannot use the quadratic formula.

Clearly $a \gt 0, b \le 0, a \gt |b|$  because the $15ab$ is negative.  We can see that $a=1,b=0$ is a solution.  If $a =1-b,$ the left side becomes $a^5+15a(1-a)-(a-1)^5=5a^4-10a^3-5a^2+10a+1$, which is greater than $1$ when $a \gt 2$ so we only need to check $a=2$  We find $a=2,b=-1$ is the only other solution.
