How to solve this Diophantus equation$(s^2=4m^2n^2+p^2$,$p^2=m^2+n^2)$?

Question

$$s^{2}=4m^{2}n^{2}+p^{2}; p^{2}=m^{2}+n^{2}; 1<m<n<p<s$$ I think that this equation does not have positive Integer solution, but how to prove?

Answer 1

HINT:

Let's assume by contradiction that there is a positive integer solution.

Hence, each of these two equations essentially yields a Pythagorean Triplet:

• The equation $p^2=m^2+n^2$ yields $[m<n<p]$
• The equation $s^2=4m^2n^2+p^2$ yields $[2mn<p<s]$ or $[p<2mn<s]$

The $[2mn<p<s]$ triplet is ruled out because:

• $m,n,p$ are the sides of a triangle $\implies p<m+n$
• $m,n$ are larger than $1 \implies m+n<2mn$

So we are left with the triplets $[m<n<p]$ and $[p<2mn<s]$.

Hence, there must be positive integers $a>b$ and $c>d$ such that:

• $m = a^2-b^2$
• $n = 2ab$
• $p = a^2+b^2$
• $p = c^2-d^2$
• $2mn = 2cd$
• $s = c^2+d^2$