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.
On
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.
Replacing $b$ with $-b$, let's rewrite the equation as
$$a^5-15ab-b^5=1$$
If $b=0$ we have $a=1$, and if $b=1$ we have $a^5=30a+2$, which implies $a=2$. So it remains to look for integer solutions with $a\gt b\ge2$. If there are any, then
$$1+15ab=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)\gt a^4+b^4\ge2(ab)^2$$
This implies $6\le ab\le 7$ (the lower bound since we're only concerned with $b\ge2$ and $a\gt b$). But $ab=7$ has no solution with $a\gt b\ge2$. We are left with the possibility $(a,b)=(3,2)$, which is easy to rule out.
Remark: This approach fairly clearly works for any equation of the form $a^n+Aab+b^n=B$ when the exponent $n$ is odd, limiting the search for solutions to a finite set of possibilities: Even after throwing away a lot of (positive) terms and ignoring multiplicative factors, you get an inequality on $ab$ that restricts it to a finite range. It's hard to throw out the baby with the bathwater here.
Of course you can hope to do better by being more circumspect. Indeed, if I'd kept the $a^2b^2$ when I threw away the $a^3b+a^2b^2+ab^3$, the inequality would have become $1+15ab\gt3(ab)^2$, which has no solutions with $6\le ab$, so there would have been no need for the final little argument. I didn't keep it for two reasons: 1) It's good to show a small example of the final step one sometimes has to take; and 2) I didn't notice I was tossing out something potentially useful.