This problem is quite well known. For example, this appeared as an example in PfTB Chapter 4. Moreover a much harder version is proposed at Iran TST 2015.

Consider prime number $p$ that $p\equiv 3\pmod{4}$. $\frac{p-1}{2}$ will be odd. If $p\mid a^2+b^2$ then, $a^2\equiv -b^2\pmod{p}$ $a^{p-1}\equiv(-1)^{\frac{p-1}{2}}b^{p-1}\pmod{p}$ $a^{p-1}\equiv -b^{p-1}\pmod{p}$ Assume that $p\nmid a$. $\Rightarrow p\nmid b$ From Fermat Little Theorem $a^{p-1}\equiv 1\pmod{p}$ and $-b^{p-1}\equiv -1\pmod{p}$. So, $1\equiv -1\pmod{p}$. $p\mid 2 \Rightarrow p=2$, a contradiction. Therefore, if $p\mid a^2+b^2$ and $p$ is prime number that $p\equiv 3\pmod{4}$ then $p\mid a$ which implies that $p\mid b$ and so $p^2\mid a^2+b^2$. $\Rightarrow$ If positive integers $x$ that $x\equiv p\pmod{p^2}$ when $p$ is prime number that $p\equiv 3\pmod{4}$ then $x$ can't be written in the form $a^2+b^2$. Now let consider $m$ such that $$m\equiv p_{1}\pmod{p_{1}^2}$$$$m\equiv p_{2}-1\pmod{p_{2}^2}$$$$m\equiv p_{3}-2\pmod{p_{3}^2}$$$$\vdots$$$$m\equiv p_{2017}-2016\pmod{p_{2017}^2}$$when $p_{1}<p_{2}<...<p_{2017}$ is different prime number such that $p_{i}\equiv 3\pmod{4}$. From Chinese Remainder Theorem There must exists $m$ that satisfies all of the above conditions. $\Box$

MarkBcc168 wrote: This problem is quite well known. For example, this appeared as an example in PfTB Chapter 4. Moreover a much harder version is proposed at Iran TST 2015. What is PfTB ?

PfTB = Problems from the Book by Titu Andreescu and Gabriel Dospinescu

Thanks !!!!!