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\[13^2=\frac{a^2}{b^2}+\frac{m^2}{n^2}+1\]

\[13^2=\frac{(pm)^2}{(qn)^2}+\frac{m^2}{n^2}+1\]\[13^2=\frac{(pm)^2+(qm)^2}{(qn)^2}+1\]

\[a,b,m,n \in \mathbb{Z}\]

Curse you, Diophantus!

:)

Are you assuming that a and b are integers?

ah! sorry - didn't read your question properly - let me think again...

Fair enough, I made the same mistake intially.