Can you write solution?

xeroxia wrote: $m < n$ are positive integers. Let $p=\frac{n^2+m^2}{\sqrt{n^2-m^2}}$. a. Find three pairs of positive integers $(m,n)$ that make $p$ prime. b. If $p$ is prime, then show that $p \equiv 1 (\mod 8)$. I may be wrong............But you may try the following. Look the denominator must be an integer. Then, obviously, $n^2-m^2=k^2$(say) $\Rightarrow n^2=k^2+m^2$ This reminds us about Pythagorean triples. Now you have to think. I am trying to find full solution.

In order for $\frac{n^2+m^2}{\sqrt{n^2-m^2}}$ to be an integer, we need $n^2-m^2$ to be a perfect square, thus $n^2 = m^2 + u^2$, and we get to use the full knowledge of Pythagorean triples. So we then need $\frac{n^2+m^2}{\sqrt{n^2-m^2}} = \frac{2m^2+u^2}{u}$, thus we need $u\mid 2m^2$. Let $d=\gcd(m,u)$, $u=da$, $m=db$, $\gcd(a,b)=1$. So we need $a\mid 2db^2$, hence $a\mid 2d$. Say $d=ka$; then $\frac{2m^2+u^2}{u} = k(2b^2 + a^2)$, and needing this to be a prime $p$ forces $k=1$ and so $p=2b^2 + a^2$. Say $a=2c$ and $d=kc$; then $\frac{2m^2+u^2}{u} = k(b^2 + 2c^2)$, and needing this to be a prime $p$ forces $k=1$ and so $p=b^2 + 2c^2$. But knowing the general parametric solution of a primitive solution for a Pythagorean triple $(u^2+v^2)^2 = (2uv)^2 + (u^2-v^2)^2$ proves that the putative prime $p$ must be $p\equiv 1 \pmod{8}$ (since then, say in $p=2b^2 + a^2$ we will have $a$ odd and $b$ even, while in $p=b^2 + 2c^2$ we will have $b$ odd and $c$ even). As for the three examples, it is enough to find primes $p=2B^2 + A^2$ that lead to a Pythagorean triple, for example $p = 2\cdot 4^2 + 3^2 = 41$, leading to $15^2 = 12^2+9^2$; $p = 2\cdot 12^2 + 5^2 = 313$, leading to $65^2 = 60^2+25^2$. For good measure, let us also use the other representation, finding a prime $p=B^2 + 2C^2$ leading to a Pythagorean triple, for example $p = 3^2 + 2\cdot 2^2= 17$, leading to $10^2 = 8^2+6^2$.

$n^2-m^2=(n+m)(n-m)$ is a perfect square. Let $n+m=a_1^2g, n-m=b_1^2g$ where $gcd(a_1, b_1)=1$ and $a_1>b_1$. So $n=\frac{(a_1^2+b_1^2)g}{2}, m=\frac{(a_1^2-b_1^2)g}{2}, n^2+m^2=\frac{(a_1^4+b_1^4)g^2}{2} \implies p=\frac{(a_1^4+b_1^4)g}{2a_1b_1}$. Since $a_1\nmid a_1^4+b_1^4$, so $a_1\mid g$. Similarly $b_1\mid g$. So $a_1b_1\mid g$. But since $p$ is prime, either Case 1: $2a_1b_1=g\implies p=a_1^4+b_1^4$ where $a_1, b_1$ have different parity. WLOG substitute $a_1=2x, b_1=2y+1$ to get $p \equiv 1 (\mod 8)$. Or Case 2: $a_1b_1=g\implies p=\frac{a_1^4+b_1^4}{2}$ where both $a_1, b_1$ are odd. Substitute $a_1=2x+1, b_1=2y+1$ to get $p \equiv 1 (\mod 8)$. To find 3 examples of $p$, let $b_1=1$ in either case. Case 1: $a_1=2, b_1=1, g=4, n=10, m=8, p=17$. Case 1: $a_1=4, b_1=1, g=8, n=68, m=60, p=257$. Case 2: $a_1=3, b_1=1, g=3, n=15, m=12, p=41$.