Prime numbers problem

Question

The natural numbers $a, b, c$ and $a^4+b^4+c^4-3$ are prime.

Prove that $a^2+b^2+ c^2-1$ is the prime number.

Answer 1

If none of $a,b,c$ is $3$

$$a^4+b^4+c^4-3\equiv0\pmod3$$

So, one of them must be $3$ let $a=3$

$$a^4+b^4+c^4-3=b^4+c^4+78$$

Now if $b,c$ both are odd, $b^4+c^4+78$ is even

$\implies b,c$ must have opposite parity

$\implies $ one of them must be $2$ let $b=2$

$$a^4+b^4+c^4-3=c^4+94$$

Now if $5\nmid c,(c,5)=1\implies c^4\equiv1\pmod5, c^4+94\equiv0\pmod5$

$\implies5\mid c\implies c=5$