Prime Sum of Squares of 3 Prime Numbers

Theorem
Let $a$, $b$ and $c$ be prime numbers with the property that:
 * $a^2 + b^2 + c^2 = p$

where $p$ is a prime number.

Then either $1$ or $2$ of $a$, $b$ and $c$ is equal to $3$.

Proof
First we note that sets of $3$ prime numbers with this property are plentiful.

For example, these are all such sets for $a, b, c < 20$:

And indeed, the number of instances of $3$ in all the above is either $1$ or $2$.

First we note that if $a = b = c = 3$ then $a^2 + b^2 + c^2 = 27$ which is not prime.

It remains to demonstrate that if none of $a$, $b$ and $c$ are equal to $3$, then $a^2 + b^2 + c^2$ is not prime.

Let $q$ be a prime number such that $q \ne 3$.

Then by definition of prime number:
 * $3 \nmid q$

where $\nmid$ denotes non-divisibility.

Then from Square Modulo 3:
 * $q^2 \pmod 3 = 1$

Suppose none of $a$, $b$ and $c$ is equal to $3$.

Then:

That is:
 * $3 \divides a^2 + b^2 + c^2$

where $\divides$ denotes divisibility.

Hence if none of $a$, $b$ and $c$ is equal to $3$, $a^2 + b^2 + c^2$ is not prime number.

The result follows.

[[Category:3]]