67th NMO Selection Tests for JBMO

a) Notice that $x=1,y=3,z=8,t=4$ is a solution to the system $(S_7)$, and so is $(k,3k,8k,4k)$, for any $k\in \mathbb{N}$, hence $7\in M_1$.

If $(x,y,z,t)$ is a solution to the system $(S_{10})$, it would follow that $11(x^2+y^2)=z^2+t^2$. From $11\mid z^2+t^2$ we get $11\mid z$ and $11\mid t$. Then $11\mid x^2+y^2$, hence $11\mid x$ and $11\mid y$. Therefore, there exist $x_1,y_1,z_1,t_1\in \mathbb{N}$ such that $x=11x_1,y=11y_1,z=11z_1,t=11t_1$. This leads to $11(x_1^2+y_1^2)=z_1^2+t_1^2$. By continuing this procedure ("infinite descent"), we get that $x$ is a multiple of any power of $11$, and, as $x\ne 0$, we arrive to a contradiction. In conclusion, the system $(S_{10})$ has no solutions, i.e. $10\in M_2$.

b) It is easy to see that any number $n$ of the form $m^2-1$, $m\in \mathbb{N}$, belongs to $M_1$: we can choose $x=y$ and notice that $(k,k,mk,mk)$, $k\in \mathbb{N}$, are solutions to the system $(S_{m^2-1})$, hence $m^2-1\in M_1,\forall m\in \mathbb{N}$. Applying the same steps as we did above for $n=10$, it is easy to prove that if $p$ is a prime of the form $4m+3$, and $n=p-1$, then $n\in M_2$. As there are infinitely many such primes, the conclusion follows readily. One can prove that actually any number congruent to $2(\mathrm{mod}4)$ belongs to $M_2$.