Let all numbers of form $x^2 +y^2$ where $x, y$ are coprime integers be arranged in a sequence $z_1 < z_2 < z_3 < . . ..$ (So the sequence begins $z_1 = 2 = 1^2 + 1^2 , z_2 = 5 = 1^2 + 2^2 , z_3 = 10 = 1^2 + 3^2$.) Prove that there exist infinitely many values of $n$ such that $z_n, z_{n+1}, …, z_{n+2019}$ are all odd.
This is from the Poland math olympiad the year $2019$.
Let $S$ be the set containing all of these numbers. Note that $z\in S$ means $z$ doesn't have a prime factor that is $\equiv 3\pmod 4$. The proof is simple, assume on the contrary that there exists $z \in S$ with $p\equiv 3\pmod 4$ being a prime factor of $z$. Then $$p\mid z=x^2+y^2\implies p\mid x \text{ and } p\mid y\implies p\mid (x,y)=1$$ A contradiction (any prime$\equiv 3\pmod 4$ has this property).
Now according to this question $z\in S \iff z$ is not divisible by $4$ and doesn't have any prime factor of the form $4k+3$. With this we have classified all the numbers in $S$.
The solution has been posted on this post on AoPS
To continue the solution outlined in the comment(which is similar to the AoPS solution, but we are constructing an $a$ such that $(a - i + 1)^2 + (a + i)^2(1 \leq i \leq 2019)$ are consecutive elements of $a_n$), you can
Take $a$ to be divisible by $10000!$, so that $(a - i + 1, a + i) = 1$ for all $1 \leq i \leq 2019$.
For each $j \in [1, 2018^2 + 2019^2]$ such that $2j + 1$ is not a square, pick a distinct prime $p_j \equiv 3 \bmod 4$ greater than $10000$ and a $b_j$ such that $p | (2b_j + 1)^2 + 2j + 1$. It is possible to show that infinitely many such $p_j$ exists as $2j + 1$ is not a square. By CRT, we can take $a$ such that $a \equiv b_j \bmod p_j$ for each $j$. This ensures that $2a^2 + 2a + j$ cannot be written as $x^2 + y^2$ for $x,y$ coprime.