Why can $5^{2n+3} + 3^{n+3} \cdot 2^{n} \quad \forall n \in \mathbb{N}$ never be a prime number?

Question

It seems to be true for the first thousand $n$ but I really can't think of a way to prove this statement.

Any kind of help will be appreciated!

Answer 1

Hint $ $ Call it $\,f_n.\, $ Note $\gcd(f_0,f_1) = 19\ $ and it's easy to verify the factor $19$ persists for all $\,n$

$\qquad\ \ {\rm mod}\,\ 19\!:\,\ f_n\equiv 5^3\, 25^n + 3^3\, 6^n \equiv 11\cdot 6^n + 27\cdot 6^n \equiv 38\cdot 6^n \equiv 0\ \Rightarrow\ 19\mid f_n$

Remark $ $ The reason that the common factor of $\,f_0,f_1\,$ persists for all $\,n\,$ is that $\,f_n\,$ satisfies a $2$nd order recurrrence $\, f_{n} = 31f_{n-1} - 150 f_{n-2}.\,$ So $\,19\mid\color{#c00}{f_0,f_1}\,\Rightarrow\, 19\mid f_2 = 31\color{#c00}{f_1} - 150 \color{#c00}{f_0}.\,$ In the same way $\,19\mid f_1,f_2\,\Rightarrow\, 19\mid f_3.\,$ This shows the way to give an alternative (strong) induction proof that $\,19\mid f_n\,$ for all $\,n\ge 0.\,$ Or you could use the recurrence and strong induction to show there are integers $\,j,k\,$ such that $\, f_{n} = j f_0 + k f_1,\,$ for all $\,n\ge 0,\,$ because such integral linear combinations of $\,f_0,f_1\,$ are closed under integer scalings and addition.

The same method works for any sequence satisfying a monic integer-coef linear recurrence.

Answer 2

Oops...I think I found a proof.

We can notice that for every $n$ our sum seems to be divisible by $19$. So we have to prove that $$5^{2n+3} + 3^{n+3} \cdot 2^{n} \equiv 0 \pmod{19}\quad \forall n \in \mathbb{N}$$ We can prove this using induction.

Base-case: \begin{align*} &19|152& \end{align*} Inductive step: \begin{align*} 0 &\equiv 5^{2(n+1)+3} + 3^{n+4} \cdot 2^{n+1} &\pmod{19} \\ 0 &\equiv 25\cdot 5^{2n+3} + 6 \cdot 3^{n+3} \cdot 2^{n} \\ 0 &\equiv 6\cdot 5^{2n+3} + 6 \cdot 3^{n+3} \cdot 2^{n} \\ 0 &\equiv 6\cdot ( 5^{2n+3} + 3^{n+3} \cdot 2^{n} )\\ 0 &\overset{\text{IH}}{\equiv} 6 \cdot 0 \equiv 0 \\&&\square \end{align*}

Answer 3

This is how I would think about it:

$(5)^{2n+3}$ is going to be an odd, since a product of all odd integers is always going to be odd (if you want me to prove that I'd be more than happy to).

So then, think of the second part of the statement as $$ A = [(2)(3)]^n \text{ and } B = (3)^3 = 27 $$ so we can write the second part as: $$ A*B $$ Since A is even, and B is odd, the product of $A$ and $B$ is going to be odd as well (again, if that doesn't makes sense, I can prove that as well).

So now we have the sum of 2 odds, which is even. And other than 2, even numbers can't be prime! Hope that helps!
-ZZ