<s>[INST] (USAJMO)Find the integer solutions:$ab^5+3=x^3,a^5b+3=y^3$ Find the integer solutions:
$$a·b^5+3=x^3,a^5·b+3=y^3$$
This is the first problem of today's USAJMO (has finished),I only find a trival result that $x\equiv y \pmod6$ and $abxy≠0 \pmod 3$.
Thanks in advance!
 [/INST] If $3 \mid a$, then $3\| (a^5b+3)=y^3$, a contradiction. Thus $3\nmid a$. Similarly $3 \nmid b$, so $3 \nmid x, y$. Note that if $3 \nmid n$, then $n^3 \equiv \pm 1 \pmod{9}$. Thus $x^3-3, y^3-3 \equiv 5, 7 \pmod{9}$, so
$$1\equiv(ab)^6 \equiv (x^3-3)(y^3-3) \equiv 4, 7, 8 \pmod{9}$$
We get a contradiction, so there are no integer solutions.
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