25. Let prime $p>2, \delta_{p}(a)=4$. Find the least positive residue of $(a+1)^{4}$ modulo $p$.


25. There must be $a^{4} \equiv 1(\bmod p), a^{2} \equiv-1(\bmod p), a^{3} \equiv-a(\bmod p)$. Therefore, $(1+a)^{4}=$ $a^{4}+4 a^{3}+6 a^{2}+4 a+1 \equiv-4(\bmod p)$. The smallest positive residue is $p-4$.