Numbers are too large to show $65^{64}+64^{65}$ is not a prime

Hint $\rm\ \ x^4 +\: 64\: y^4\ =\ (x^2+ 8\:y^2)^2 - (4xy)^2\ =\ (x^2-4xy + 8y^2)\:(x^2+4xy+8y^2)$

Thus $\rm\ x^{64} + 64\: y^{64} =\ (x^{32} - 4 x^{16} y^{16} + 8 y^{32})\:(x^{32} - 4 x^{16} y^{16} + 8 y^{32})$

Below are some other factorizations which frequently prove useful for integer factorization. Aurifeuille, Le Lasseur and Lucas discovered so-called Aurifeuillian factorizations of cyclotomic polynomials $\rm\;\Phi_n(x) = C_n(x)^2 - n\ x\ D_n(x)^2\,$ (aka Aurifeuillean). These play a role in factoring numbers of the form $\rm\; b^n \pm 1\:$, cf. the Cunningham Project. Below are some simple examples of such factorizations:

$$\begin{array}{rl} x^4 + 2^2 \quad=& (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\\\ \frac{x^6 + 3^2}{x^2 + 3} \quad=& (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\\\ \frac{x^{10} - 5^5}{x^2 - 5} \quad=& (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\\\ \frac{x^{12} + 6^6}{x^4 + 36} \quad=& (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\\\ \end{array}$$

$$64^{65}+65^{64} = 6^{65}+7^{64} \pmod{29}$$

$65=2 \times 28+9, 64 = 2 \times 28 +8$, and also gcd$(29,36)$ = gcd$(29,49) = 1$

Therefore by Fermat's Little Theorem

If gcd$(a,p)= 1$, and $p$ is a prime then $a^{(p-1)} \hspace{3pt}\equiv \hspace{3pt}1 \pmod{p}$

$36^{29-1} \equiv 1 \pmod{29}, \hspace{5pt}49^{29-1} \equiv 1 \pmod{29} \hspace{3pt} \implies \hspace{3pt} (6^{2})^{28} \equiv 1 \pmod{29}, \hspace{5pt} (7^{2})^{28} \equiv 1 \pmod{29}$

Therefore $6^{65} = 6^{(56+9)} \equiv 6^9 \pmod{29}, \hspace{5pt} 7^{64} = 7^{(56+8)} \equiv 7^8 \pmod{29}$

$$64^{65}+65^{64} \equiv 6^9+7^8 \pmod{29} \hspace{5pt} \equiv 22+7 \pmod{29} \equiv 0 \pmod{29}$$

Which shows that $$29 | (65^{64}+64^{65})$$

Therefore $65^{64}+64^{65}$ is not a prime.

Hint: use the Sophie Germain identity which states

$$x^4+4y^4=(x^2+2xy+2y^2)(x^2-2xy+2y^2)$$