if $m,n\geq 1$ if $\frac{1+p^2+q^2}{pq} \in \mathbb{N}$ it is $3$ so $p^{m-1}q^{n-1}=5$ and we find the solutions (m,n,p,q)=(2,2,5,1) and (2,1,5,13) (2,1,5,2) if m=0 then $q^n=(p+q)^2+1$ which is solvable.

Albanian Eagle wrote: $q^n=(p+q)^2+1$ which is solvable. Can you show me a solution of this part?,I have found a sloution,but I used the expretion$a^2+1=(a-i)(a+i)$ are ther other solutions,without complex numbers ?

The main ideea is that the ecuation $q^n=(p+q)^2+1$ is quadratic in p, an if it has solution p it has also $2q-p$ because forom Viete the sum of roots is $2q$. Hence we can work with that root $\le q$. For big enough values of $p,q,n$ (greater than 3 or 4 is enough) we can prove that $q^n > (p+q)^2+1$. The details can be filled easily.

xirti wrote: The main ideea is that the ecuation $q^n=(p+q)^2+1$ is quadratic in p, an if it has solution p it has also $2q-p$ because forom Viete the sum of roots is $2q$. Hence we can work with that root $\le q$. For big enough values of $p,q,n$ (greater than 3 or 4 is enough) we can prove that $q^n > (p+q)^2+1$. The details can be filled easily. sum of roots is -2q<0.

Albanian Eagle wrote: if $\frac{1+p^2+q^2}{pq} \in \mathbb{N}$ it is $3$ can you show that please.

of course show by infinite descent that if $p^2+ q^2 = apq-1$ then a must be $3$

Albanian Eagle wrote: of course show by infinite descent that if $p^2+ q^2 = apq-1$ then a must be $3$ You havn't answered my question about second part!

Tiks wrote: Albanian Eagle wrote: of course show by infinite descent that if $p^{2}+q^{2}= apq-1$ then a must be $3$ You havn't answered my question about second part! Yeah...

But I thought that's pretty classic

You Can Also See The Answer In The Following Link : http://www.irysc.com/forum/t6886-post90551/#post90551 Post #8 Is An Answer And Post #11 Is Another Answer To This Question.

I find a solution without using the fact: $\frac{p^2+q^2+1}{pq}=3$, which of course solves easily the problem. We see that $(p+q)^2+1\ge 1$, so $p,q\ne 0$. First, if $m=0$ or $n=0$ we get the following equations $q^n=(p+q)^2+1$ and $p^m=(p+q)^2+1$, respectively. But for a result of V. Lebesgue http://www.math.leidenuniv.nl/~jdaems/scriptie/Catalan.pdf we know these equations have no non-zero solutions. Therefore we may assume that $m,n\ge 1$. So $m+n\ge 2$. If $m+n=2$ we get the equation $pq=(p+q)^2+1$ and it's easy to see that there are no solutions for $p,q>0$. Then $m+n>2$. We have $p^{m}q^{n}=p^2+2pq+q^2+1$, so we infer that $p\mid q^2+1$ and $q\mid p^2+1$, and also if $(m,n,p,q)$ is a solution then $(n,m,q,p)$ is a solution too. Then we assume without loss of generality that $p<q$ (it's clear that $p=q$ is impossible). If $q=p^2+1$ we would have $p\mid 2$, so we found the pair $(2,5)$. Replacing into the equation we get $2^{m}5^{n}=50$, hence $m=1$ and $n=2$. Then we have the quadruple $(1,2,2,5)$. Now we assume that $q\ne p^2+1$. So $q\le \frac{p^2+1}{2}$. We check for $p\le 4$ if there are solutions and we don't find a new pair apart from $(2,5)$. From now on we suppose that $q>p\ge 5$. We have $p^{m+n}<p^{m}q^{n}=(p+q)^2+1\le (p+\frac{p^2+1}{2})^2+1=\frac{(p+1)^4}{4}+1$, so $4p^{m+n}\le (p+1)^4$. If $m+n\ge 4$ we have $4p^4\le (p+1)^4$ but $3p^4>4p^3+6p^2+4p+1$ for all $p\ge 5$, hence $m+n<4$. So we have found $2<m+n<4$. Therefore $m+n=3$. On the other hand $5q^n\le p^{m}q^{n}=(p+q)^2+1<(2q)^2+1$, then $5q^n\le 4q^2$, if $n\ge 2$ we get $5q^2\le 4q^2$, impossible because $q>0$, hence $n=1$ and so $m=2$. Finally we have the equation $p^{2}q=(p+q)^2+1=p^2+2pq+q^2+1$. If we see this equation as a quadratic equation in $q$ we get $q^2+(2p-p^2)q+(p^2+1)=0$. In order to have $q\in \mathbb{Z}$ the discriminant of last equation must be a perfect square, so $(2p-p^2)^2-4(p^2+1)=n^2$. Therefore $p^4-4p^3-4=n^2$. But we have $p^4-4p^3-4>(p^2-2p-5)^2$ and $p^4-4p^3-4<(p^2-2p-2)^2$ for $p\ge 5$ (the reader can check these inequalities). Then $n=p^2-2p-3$ or $n=p^2-2p-4$. In the first case we get the equation $2p^2-12p-13=0$ and this equation doesn't have integer solutions. In the second case we get $p^2-4p-5$, so we find $p=-1$ and $p=5$. Since $p=5$ is only valid we get $q^2-15q+26=0$, then we find $q=2$ and $q=13$. Since $q>p$ we have $q=13$. Therefore we have the quadruple $(2,1,5,13)$. Hence all the solutions are $(1,2,2,5), (2,1,5,2), (2,1,5,13)$ and $(1,2,13,5)$. I also comment that there is another solution for the equation $p^{2}q=(p+q)^2+1$ if we see it as a quadratic equation in $p$.