Can we prove that $26^n$ always ends in the same two digits?

Can we prove that $26^n$ always ends in the same two digits?

Question

Prove that, for all positive integral values of $m$, $x-1$ is a factor of $x^m - 1$.

Prove that, if $n$ is an integer greater than $1$, then $26^n - 26$ is an odd multiple of $50$.

Hence, or otherwise, show that, if $26^n$ is evaluated for any integral value of $n$ greater than $1$, then the last two digits of the result are independent of $n$.

[The phrase ‘integral values’ means the same as ‘integer values’. ]

Solution

By the factor theorem, a polynomial $p(x)$ has a linear factor $(x-a)$ if and only if $p(a) = 0$.

If we take $m$ to be a positive integer and set $p(x) = x^m - 1$, then $p(1) = 1^m - 1 = 0$ and therefore $(x-1)$ is a factor of $x^m - 1$.

Moreover, we can write $x^m - 1 = (x-1)(x^{m-1} + \dots b + 1).$

As

$n > 1$

is an integer, we can write (using the first part of the question)

$$\begin{align*} 26^n - 26 &= 26 \times ( 26^{n-1} - 1 ) \\ &= 26 \times (26 - 1) \times (26^{n-2} + \dots b +26^1 +1) \\ &= 50 \times 13 \times (26^{n-2} + \dots b + 26^1 + 1). \end{align*}$$

Since $26^{n-2} + \dots b + 26^1$ is an even integer, the bracketed factor is odd, and so $13$ times the bracketed factor is also odd. That is, $26^n - 26$ is an odd multiple of $50$.

We can write $26^n = \left( 26^n - 26 \right) + 26.$ By the second part of the question (which applies as $n > 1$ is an integer), as the bracketed term is an odd multiple of $50$, its final two digits will always be $50$.

Thus, the final two digits of the right-hand side of the above will be $76$; this is independent of the value of $n$.