Remainder of $1946^{1972} : 26$

Question

Is this correct?

$1946^1 = 22 \mod{26}$

$1946^2 = 22^2 = 484 = 16\mod{26}$

$1946^3 = 22^2 * 22 = 16 * 22 = 14 \mod{26}$

$1946^4 = 22^2 * 22^2 = 16^2 = 22 \mod{26}$

And therefore for any integer $k \geq 0$:

$1946^{3k+1} = 22 \mod{26}$

$1946^{3k +2} = 16 \mod{26}$

$1946^{3k+3} = 14 \mod{26}$

Then: $1972 = 3*657 +1 \rightarrow k = 657$

And the answer is: $1946^{3*657+1} = 22 \mod{26}$

---

How can I use the same algorithm for $(1946^{1972} + 1972^{1946}) \mod{26}$ ?

Answer 1

Hint:

$1946 \equiv 1972 \pmod{26}$, because $1946 + 26 = 1972$. You can reuse your table of powers of $1946 \pmod{26}$ to compute the desired power of $1972 \pmod{26}$.

Then you can easily perform the addition and the last mod step.