solution verification | Determine the prime natural numbers $p$ for which there exists $x \in Z$ such that $p | x^2+2x+1$ and $p | x^2+8x+16$.

Question

The questions

Determine the prime natural numbers $p$ for which there exists $x \in Z$ such that $p | x^2+2x+1$ and $p | x^2+8x+16$.

my idea

We know that

$x^2+2x+1=(x+1)^2$

$x^2+8x+16=(x+4)^2$

which means

$$p | (x+1)^2$$

and

$$p | (x+4)^2$$

because p is prime it means that

$p | (x+1)$ and $p | (x+4)$

if we decfrease them we get that $p|3$ and because p is natural we get that p can only be 3

I'm not sure about the part where i said that because p is prime it means that

$p | (x+1)$ and $p | (x+4)$

I feel like i should explain more, but i dont know how to. Hope one of you can tell me if my rationament is good and how should i improve the answer. Thank you!

Yes, if follows immediately by Euclid's Lemma as you wrote. For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. — Bill Dubuque, Feb 12 '24 at 16:31

score 1 · Accepted Answer · answered Feb 12 '24 at 15:39

Your analysis is both accurate and valid.

If $~p~|~n^2,~$ then $~p~$ occurs in the prime factorization of $~n^2.~$

Since the prime factorization of $~n~$ is identical to the prime factorization of $~n^2,~$ except that the exponents are different, $~p~$ must also occur in the prime factorization of $~n.~$

This implies that $~p ~| ~n.$

solution verification | Determine the prime natural numbers $p$ for which there exists $x \in Z$ such that $p | x^2+2x+1$ and $p | x^2+8x+16$.

1 Answers1