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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!

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up vote 7 down vote accepted

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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($+1$) Credits to you, sir. Helping SE with unanswered questions –  Inceptio May 1 '13 at 15:01
    
Without the information supplied by hecke, why is the fact that $3|y^3$ a contradiction? –  chubakueno Jul 20 '13 at 1:58
    
I wrote that $3 \| y^3$, not $3|y^3$. The latter indeed does not give an immediate contradiction, but the former clearly does. –  Ivan Loh Jul 28 '13 at 23:23
    
To summarize: there are no solutions mod 9. –  zyx Apr 1 at 0:05
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