Call Us 07766496223
Let  
\[p\]
  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  
\[1^1+(-4)^2=17\]
.