\[p\]
be a prime number of the form \[p=4k+1\]
and let \[x, \; y\]
be the least absolute residues modulo \[p\]
\[x \equiv \frac{(2k)!}{2(k!)^2} \; (mod \; p), \; \; y \equiv (2k)!x \; (mod \; p)\]
.Then
\[17=4 \times 4+1\]
.\[\]
is prime.\[x \equiv \frac{(2 \times 4)!}{2 \times (4!)^2}=35 \; (mod \; 17) \equiv 1 \; (mod \; 17) \]
\[y\equiv (2 \times 8 )! \times 1 \; (mod \; 17) \equiv - 4 \; (mod \; 17)\]
Then
\[1^1+(-4)^2=17\]
.