Super easy problem Sum of two squares, p=4k+3 gg

Indeed, $f(x) = (x^2+x-2)^2 + x^2$, and $\text{gcd}(x^2+x-2,x) = \text{gcd}(x,2)$.

Here's an overkill solution: Lemma 1. Let $P \in \mathbb{Z}[x]$ be irreducible (in $\mathbb{Q}$) and of degree $\ge 2$. Then there exists infinitely many primes $p$ such that $p$ does not divide $P(m)$ for any $m \in \mathbb{Z}$. Lemma 1 is a quite well-known application of Chebotarev density theorem. And now we claim that $f$ is irreducible. This can be handled by the following quite interesting lemma: Lemma 2. Let $P \in \mathbb{Z}[x]$. $P$ is reducible in $\mathbb{Z}$ if (and only if) $P$ is reducible in $\mathbb{Q}$. Proof. This has been proved in https://math.dartmouth.edu/archive/m31w05/public_html/Q-Z-Reducibility.pdf . The main idea is to assume $P = QR$, with $Q, R \in \mathbb{Q}[x]$. There exists an integer $c > 0$ such that $cP = Q'R'$, $Q, R' \in \mathbb{Z}[x]$. If $c = 1$, we are done, and otherwise we pick a prime divisor $p$ of $c$. Looking at the equation $cP = Q'R'$ modulo $p$ it is not hard to see that either $P'$ or $Q'$ has all of it's coefficients divisible by $p$. We can divide the equation by $p$. We can apply this process until $c = 1$, which proves the claim. I am not sure if lemma 2 is well-known. I just learned about it a couple days ago, which is the reason I wanted to share it. I think I've heard of a proof for monic polynomials $P$, but the general result was a new one for me. It remains to prove that there are no $Q, R \in \mathbb{Z}[x]$ such that $f = QR$. It is easy to see that $Q$ and $R$ have degree $2$, as $f$ has no rational roots. WLOG. $Q(x) = x^2 + ax + b$ and $R(x) = x^2 + cx + d$. We now have $bd = 4$, which means we only have a couple options for $b, d$. Comparing the coefficients of $f$ and $QR$ we can derive the equations $a + c = 2$ and $b + d + ac = -2$. If we know $b, d$ it is easy to calculate $a, c$. Details are left to the reader.

Loppukilpailija wrote: Lemma 2. Let $P \in \mathbb{Z}[x]$. $P$ is reducible in $\mathbb{Z}$ if (and only if) $P$ is reducible in $\mathbb{Q}$. This is called Gauss lemma. https://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomial)

I can't know that this is problem3. I solved this by 10 minutes!

This is actually easiest problem ever.. I solve this in 5 minute as well