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The question is equivalent to the system of quadratic Diophantine equations: $$8 n^2 = m'^2 - 1$$ $$4n^2 = 3p'^2 + 1$$ where $m'=2m+1$ and $p'=2p+1$. How to solve such systems is described in my paper: http://arxiv.org/abs/1002.1679 (see Theorem 6) It is easy to obtain that the only nonegative solution is $n=1$, $m=1$, $p=0$. |
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The question is equivalent to the system of quadratic equations: $$8 n^2 = m'^2 - 1$$ $$4n^2 = 3p'^2 + 1$$ where $m'=2m+1$ and $p'=2p+1$. How to solve such systems is described in my paper: http://arxiv.org/abs/1002.1679 (see Theorem 56) It is easy to obtain that the only nonegative solution is $n=1$, $m=1$, $p=0$. |
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The question is equivalent to the system of quadratic equations: $$8 n^2 = m'^2 - 1$$ $$4n^2 = 3p'^2 + 1$$ where $m'=2m+1$ and $p'=2p+1$. How to solve such systems is described in my paper: http://arxiv.org/abs/1002.1679 (see Theorem 5) It is easy to obtain that the only nonegative solution is $n=1$, $m=1$, $p=0$. |
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