16th Junior Turkish Mathematical Olympiad

a. For $(m,n)=(6,10)$, $(12,15)$ and $(30,78)$, $p$ is equal to $$\frac{10^2+6^2}{8}=17,\frac{15^2+12^2}{9}=41\text{ and }\frac{78^2+30^2}{72}=97.$$ respectively.

b. Let $k=\sqrt{n^2-m^2}$. Then $(k,m,n)$ is a Pythagorean triple. Therefore, there exist positive integers $d,x$ and $y$ such that $(x,y)=1$, $n=d(x^2+y^2)$ and $m=d(x^2-y^2)$ or $m=2dxy$.

If $m=d(x^2-y^2)$, then $p=\frac{d(x^4+y^4)}{xy}$. Since $(x,y)=1$, we have $(xy,x^4+y^4)=1$ and hence $xy$ divides $d$. As $x^4+y^4\ge 17$ and $p$ is a prime, we get $d=xy$ and $p=x^4+y^4$. Since one of $x$ and $y$ is even and the other one is odd, $p=x^4+y^4\equiv 1(\mathrm{mod}8)$.

If $m=2dxy$, then $p=d(x^2-y^2)+\frac{8d{x}^2{y}^2}{x^2-y^2}$. Since $\frac{8d{x}^2{y}^2}{x^2-y^2}$ is an integer and $(xy,x^2-y^2)=1$, we have $x^2-y^2$ divides $8d$. On the other hand, as $p$ is a prime number, $d$ and $\frac{8d}{x^2-y^2}$ are relatively prime. Thus, $x^2-y^2$ is a multiple of $d$. Therefore, $x^2-y^2=d,2d,4d$ or $8d$. If $x^2-y^2=d$, then as $p=d^2+8x^2{y}^2$ is a prime, $d$ is odd and $p\equiv 1(\mathrm{mod}8)$. If $x^2-y^2=2d$, then $p=2(d^2+2x^2{y}^2)$ is not a prime number. If $x^2-y^2=4d$, then $p=2(2d^2+x^2{y}^2)$ is not a prime number.

If $x^2-y^2=8d$, then $p=8d^2+x^2{y}^2$ is a prime number implies that both $x$ and $y$ are odd numbers and consequently $p\equiv 1(\mathrm{mod}8)$.