Looks small

Solving the given modular equation is equivalent to solving the equation $p^2+1\equiv 25p \mod 143$ Now, note that, we can reduce the given equation to the following: $11(p^2+p+1)\equiv 0\mod 143$ Since $143=11\times 13$ , it follows that $p^2+p+1\equiv 0\mod 13$ . This implies that $p\equiv 3$ , or $p\equiv 9\ \mod 13$ . Now, if $p= 13k+3$ , from the equivalent of the given condition, we find, $13k^2-19k-5\equiv 0\mod 11\implies k^2+(k-4)^2\equiv 10\mod 11\implies k\equiv 1,3\mod 11$ . Similarly, if $p=13k+9$ , we find, $13k^2-7k-11\equiv 0\mod 11\implies 2k^2-7k\equiv 0\mod 11\implies k\equiv 0,9\mod 11$ . Thus, we find that $p\equiv 13k+a\equiv 9,16,42,126\mod 143$ . An easy search then reveals that $269$ is the smallest prime number of this form. Thus the answer is $\boxed{269}$ .