find two primes less than $500$ that divide $20^{22}+1$ 
Find two primes less than $500$ that divide $20^{22}+1$.

Note that $20^2+1=401$ divides the required number (since for any integer a, if k is odd, then $a+1$ divides $a^k+1$). Also, any prime dividing the given number must be congruent to $1$ modulo $4$ since $-1$ would be a quadratic residue modulo any such prime. Primes like $17$ and $13$ don't work. $20\equiv 7\mod 13, 20\equiv 3\mod 17,$ and in both cases $20$ is a primitive root with respect to the corresponding modulus. So it might be that we'll need to find a large prime. $a^k+1 = (a+1)(a^{(k-1)} - a^{(k-2)}+\cdots + 1),$ where $a = 20^2, k=11.$  I'm not sure if there's some way to factor $a^{(k-1)} -a^{(k-2)}+\cdots + 1.$

I claim that the two primes are $89$ and $401$. The $401$ is kind of obvious because $20^{22}=(20^2)^{11}\equiv (-1)^{11}\equiv -1\pmod{401}$. The $89$ is a bit more subtle. Note that $2^{44}\pmod{89}$ is just $\left(\frac{2}{89}\right)$, the Legendre symbol. It's easy to calculate by quadratic reciprocity that this is $1$. I'm too lazy to compute $5^{22}\pmod{89}$ the smart way, so just reduce: $5^{22}\equiv 5\cdot 125^7\equiv 5\cdot 36^7\equiv 5\cdot 36\cdot 1296^3\equiv 2\cdot 50^3\equiv -1\pmod{89}$. Then $20^{22}\equiv 5^{22}\cdot 2^{44}\equiv -1\pmod{89}$.
