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show that this diophantine equation: $$b^2=a^3+ac^4$$ has no soluton in non-zero integers

[Hint: first show that $a$ must be a perfect square]

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I know this reslut$$a^4+b^4=c^2$$ have no solution in non-zero integers, so this problem is key prove $a$ is perfect square.

My idea: since $$b^2=a(a^2+c^4)$$ it is easy when $a=1,a^2+c^4=1+c^4=b^2$ have no integer nozero solution so let $$a=bk_{1},a^2+c^4=bk_{2}?$$ then I can't

math110
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  • The right equations are $a=db_1^2,\ a^2+c^4=db_2^2,$ where $d=\gcd(a,a^2+c^4).$ – 111 Jan 15 '15 at 13:53

1 Answers1

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Suppose, $p$ is a prime, such that $p^u$ is the largest power dividing $a$ and the exponent is odd.

If $p$ does not divide $c$, the largest power $p^u$ dividing $a(a^2+c^4)$ is $p^u$, so the exponent is also odd.

If $p$ divides $c$, the $v$ in the largest power $p^v$ dividing $a^2+c^4$ is even. $p^{u+v}$ is the largest power of $p$ dividing $a(a^2+c^4)$ , and $v+u$ is odd.

So, $a(a^2+c^4)$ cannot be a perfect square in any case.

Peter
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