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Frobenius and Hurwitz( in 1880) prove this theorem:

For any positive integer $k$ other than 1 or 3, the equation $a^2+b^2+c^2=kabc$ has no integral solution except (0,0,0).

My Question,How to solve this following equation postive integer solutions $$a^2+b^2+c^2+1=kabc$$

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    $a^2+b^2+c^2=abc$ resp. $a^2+b^2+c^2=3abc$ have infinitely many solutions, by the way. So what do you assume on $k$ for the "modified" Markov equation ? For $k=4$ we obtain another Markov equation, see here. – Dietrich Burde Jun 09 '15 at 18:22

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You could consider the general form of this equation

$ a^2+b^2+c^2+d^2=kabcd $

Note that if $(a,b,c,d)$ is a solution then so is $(kbcd-a,b,c,d)$.

We observe that $1^2+1^2+1^2+1^2=4.1.1.1.1$ so using the starting solution $(1,1,1,1)$ we can generate an infinite number of solutions using the recurrence relation for k=4.

dear chap
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