Find all triplets $(m, n, p)$, where $p$ is a prime number and $m, n ∈ \Bbb N$, such that
$p=\frac{m}{4}\sqrt{{2n-m \over 2n+m}}$
My procedure is as follows:
$p=\frac{m}{4}{\sqrt{(2n)^2-m^2}\over 2n+m}$
It can be shown that m cannot be odd, so if $m=2k$,
$p=\frac{k}{2}{\sqrt{n^2-k^2}\over n+k}$
If $l=\sqrt{n^2-k^2}$ then $(k,l,n)$ form a Pythagorean triplet giving the two equations
$p=\frac{kl}{2(n+k)}, k^2+l^2=n^2$
I do not know how to proceed after this. Using a python script, I think the only solutions are $(24,15,2),(24,20,3)$ and $(30,39,5)$. How do I prove this?
Any help would be appreciated, thanks.
Let $m$ and $n$ be positive integers, and $p$ a prime number, such that $$p=\frac{m}{4}\sqrt{\frac{2n-m}{2n+m}}.$$ Then squaring and clearing denominators yields $$16p^2(2n+m)=m^2(2n-m).$$ Note that $m$ is even, say $m=2k$ for some positive integer $k$, and so $$4p^2(n+k)=k^2(n-k).$$ Let $d:=\gcd(k,n)$ so that $k=da$ and $n=db$ with $\gcd(a,b)=1$. Then also $$4p^2(b+a)=d^2a^2(b-a).\tag{1}$$ Of course $a^2$ is coprime to $b+a$ and so $a^2$ divides $4p^2$, meaning that $a$ divides $2p$. That is to say $a\in\{1,2,p,2p\}$.
If $a=1$ then equation $(1)$ reduces to $$4p^2(b+1)=d^2(b-1).$$ Of course $\gcd(b+1,b-1)$ divides $2$, so they are either both perfect squares, or both twice perfect squares. The former is impossible as no two perfect squares differ by $2$. In the latter case we easily find that $b=1$, but then the right hand side vanishes, a contradiction.
If $a=2$ then equation $(1)$ reduces to $$p^2(b+2)=d^2(b-2).$$ Now $\gcd(b+2,b-2)=1$ because $b$ is odd, and so $b+2$ and $b-2$ are both perfect squares. As before this again easily implies that $b=2$, but then the right hand side vanishes again, a contradiction.
If $a=p$ then equation $(1)$ reduces to $$4(b+p)=d^2(b-p),$$ where $p$ does not divide $b$ because $\gcd(a,b)=1$. We see that $d>2$, and we can rearrange the above to $$(d^2+4)p=(d^2-4)b,$$ where $p$ divides $d^2-4=(d+2)(d-2)$. If $p$ divides $d+2$ then $d-2$ divides $d^2+4$, and hence also $$d^2+4=(d-2)(d+2)+8,$$ and so $d\in\{1,3,4,6,10\}$. If $p$ divides $d-2$ then similarly $d+2$ divides $8$ and so $d\in\{2,6\}$. Either way we see that $d\in\{3,4,6,10\}$, as we already noted that $d>2$. Correspondingly we get \begin{eqnarray} 13p&=&5b\qquad\text{ and so }\qquad (m,n,p)&=&(30,39,5),\\ 20p&=&12b\qquad\text{ and so }\qquad (m,n,p)&=&(24,20,3)\\ 40p&=&32b,&&\\ 104p&=&96b,&& \end{eqnarray} and the latter two do not have solutions with $p$ prime.
If $a=2p$ then equation $(1)$ reduces to $$b+2p=d^2(b-2p).$$ Note that $b$ is odd because $\gcd(a,b)=1$, hence also $d$ is odd. We can rearrange the above to $$2(d^2+1)p=(d^2-1)b.$$ The right hand side is divisible by $8$ because $d$ is odd. The left hand side is then also divisible by $8$, which forces $p=2$. Then we are left with $$b+4=(b-4)d^2,$$ from which we easily conclude that $b=5$ and $d=3$, yielding $(m,n,p)=(24,15,2)$.