Task: Prove
Prove that for all x $\in \mathbb{N}$ it holds that the number a(n) := $x^2$ + 4x + 3 has a non-trivial divisor.
Non-trivial divisor: A non-trivial divisor of the number a is a natural number n with n!= 1 and n != a, so that a is multiple of n.
I tried to prove it with induction, but I am not quite able to finish the proof.
What I have so far:
Base case: a(1) = 2 + 4 + 3 = 8; 8 = 4 * 2, so this holds
Hypothesis: a(x) = $x^2$ + 4x + 3 = y * n with n != 1 and n != a holds
Now we have to show that a(x+1) holds:
a(x+1) = $(x+1)^2$ + 4(x+1) + 3 = $x^2$ + 2x + 1 + 4x + 4 + 3 =
= $x^2$ + 6x + 8
Now I can write: $x^2$ + 6x + 8 = ($x^2$ + 4x + 3) + (2x +5)
and substitute ($x^2$ + 4x + 3) with y * n from the hypothesis
Then I get: y * x + 2x + 5
And here I am stuck, I do not know how to continue.
Any help would be really appreciated.
Hint: $$a_n:= n^2 + 4n + 3 = (n+1)(n+3)\quad \text{ for }\quad n\in \mathbb N, n\geq 1\tag 1$$
So $$a_1 = (1+1)(1+3) = 2\cdot 4$$
Note also that $$n^2 + 6x + 8 = (n+2)(n + 4) = ((n + 1)+1)((n+1) + 3)= a_{n+1}$$
In the end, proof by induction can work, but it is not necessary. Noting $(1)$ above, given any $n \in \mathbb N, n\ge 1$, $\quad a_n= n^2 + 4n + 3 = (n+1)(n+3)$ is, by definition, divisible by $n+1,$ and also by $n+3,$ each of which are non-trivial factors of $a_n$