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Solving $a^5=a^3bc+b^2c$ in integers.

I tried assuming there is a common divisor first of a,b,c, then a,b and 2 other pairs, but not sure how to arrive to a contradiction, trying some things right now

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1 Answer 1

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Hint:

$cb^2+a^3cb-a^5=0 \Rightarrow a^6c^2+4a^5c$ is a square number, so $a^2c^2+4ac$ should be square number.

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  • $\begingroup$ Yes, that's then a necessary condition. Does it lead to a contradiction? $\endgroup$
    – coffeemath
    Commented Jan 9, 2015 at 5:38
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    $\begingroup$ @coffeemath that makes $a^2c^2+4ac$ and $(ac+2)^2$ two perfect squares with difference $4$, we know there are only a finitely many pairs of perfect squares that differ by $4$ :-) $\endgroup$
    – r9m
    Commented Jan 9, 2015 at 6:04
  • $\begingroup$ this is brilliant! :)how come i didn't notice the quadratic trinomial $\endgroup$ Commented Jan 9, 2015 at 6:23
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    $\begingroup$ @r9m I see your comment, and checked -- the squares would have to be $0,4$ and no solution. (+1 on comment, and on Belanov's answer.) $\endgroup$
    – coffeemath
    Commented Jan 9, 2015 at 6:55

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