Find the largest prime!

$5^{2n+1}+2^{n+4}+2^{n+1}=5*25^n+18*2^n$

For $n=1$ , the value of expression is equal to $161=7*23$ . So the answer is $0, 7$ or $23$ . If we prove that it is always divisible by $23$ , then the answer is $23$ .

$\begin{aligned} 5^{2n+1}+2^{n+4}+2^{n+1} & =5*25^n+18*2^n \\ & = 5*25^n-5*2^n+5*2^n+18*2^n \\ & = 5*(25^n-2^n)+23*2^n \end{aligned}$

Since $a-b\mid a^n-b^n$ , where $a, b, n$ are three positive integers, $5*(25^n-2^n)+23*2^n$ is divisible $25-2=23$ .

Therefore the answer is $\boxed{23}$ .

Rearrange the expression like this:

5 * 5^(2n) + 16 * 2^n + 2 * 2^n

Then reduce the power of the 5 term to get:

5 * 25^n + 16 * 2^n + 2 * 2^n

Then add like terms to get:

5 * 25^n + 18 * 2^n

Now, it's good to note that for n=1 and n=2, this expression is divisible by 23. But we want to check it for all n, so let's take it mod 23 (kind of). Note that 25 and 2 are both congruent to 2 (mod 23). That means:

5 * 25^n + 18 * 2^n

is congruent to:

5 * 2^n + 18 * 2^n

(mod 23).

When you add like terms, you get 23 * 2^n, which is obviously congruent to 0 (mod 23). So that proves that the expression is always divisible by 23 .

Note: This probably looks like a mess to you. I really need to learn how to use LaTeX.

Taking n=0 gives the result of 23, which is prime. So p=23 or p=1, but the latter is not prime, so p must be 23. There's 1 problem though... If you set n=1 you get 163 which isnt divisible by 23. So it doesn't hold for any n.