<s>[INST] Show that $p$ is a divisor of numbers $m^2 + 1$, $n^2 + 2$, if and only if it is a divisor of some number of the form $k^4 + 1$. Show that $p$ is a divisor of numbers  $m^2 + 1$, $n^2 + 2$, if and only if it is a divisor of some number of the form $k^4 + 1$.
Since $p \mid m^2 + 1$, we have $m^2 \cong -1( \mod p)$ and hence $\left ( \frac{-1}{p} \right)$ (Legendre Symbol) is $1$.
Again, $p \mid n^2 + 2$, we have $n^2 \cong -2( \mod p)$ and hence $\left ( \frac{-2}{p} \right)$ (Legendre Symbol) is $1$.
I am stuck here...How to proceed with the problem...Help Needed.
Thank You.
 [/INST] We have in $\Bbb F_p$
$$x^2=-1\iff (-1)^{\frac{p-1}{2}}=1\Rightarrow p=4s+1\\x^2=-2\iff (-2)^{\frac{p-1}{2}}=1\Rightarrow p=8t\pm 1$$
Hence we must have $p=8k+1$ (the cases $p=8t-1$ must be discarded)
On the other hand $$(-1)^{\frac{p-1}{2}}=((-1)^{\frac{p-1}{4}})^2$$
This means, since $\frac{8t+1-1}{4}$ is a natural integer, that there exists $k\in\Bbb F_p$ such that $$k^4+1=0\iff k^4+1=Mp$$   
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