Is sum of square of primes a square of prime? I would like to know if it has been proved that :


*

*There are no $a$, $b$ and $c$, all prime numbers, such that $a^2 + b^2 = c^2$

*There are no $a$, $b$, $c$ and $d$, all prime numbers, such that $a^2 + b^2 + c^2 = d^2$

*There are no $a$, $b$, $c$, $d$ and $e$, all prime numbers, such that $a^2 + b^2 + c^2 + d^2 = e^2$


Thanks for your answers.

It turns out that the (full) solution is longer than a comment. So here it goes.

Suppose that for $a,b,c,d$ primes,
$$a^2+b^2+c^2=d^2$$
Obviously $d=3$ is not a solution. That implies
$$a^2+b^2+c^2\equiv 1\pmod{3}.$$
So exact two of $a,b,c$ are equal to $3$. We are left to solve
$$d^2-a^2=(d-a)(d+a)=18.$$
But $d-a$ and $d+a$ are of the same parity and $18$ cannot be factorized into such a product.

For the third equations, note that $n^2\equiv 1\pmod{4}$ if $n$ is odd. Thus if 
$$a^2+b^2+c^2+d^2=e^2,$$
then exactly $3$ of $a,b,c,d$ are $2$. We are left to solve
$$(e-d)(e+d)=12=6\cdot 2.$$
It follows that $e+d=6$, and $e-d=2$. That means $e=4$, not a prime.
