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 $a, b, c < 20$:
| \(\ds 2^2 + 2^2 + 3^2\) | \(=\) | \(\ds 17\) | ||||||||||||
| \(\ds 3^2 + 3^2 + 5^2\) | \(=\) | \(\ds 43\) | ||||||||||||
| \(\ds 3^2 + 3^2 + 7^2\) | \(=\) | \(\ds 67\) | ||||||||||||
| \(\ds 3^2 + 3^2 + 11^2\) | \(=\) | \(\ds 139\) | ||||||||||||
| \(\ds 3^2 + 3^2 + 17^2\) | \(=\) | \(\ds 307\) | ||||||||||||
| \(\ds 3^2 + 3^2 + 19^2\) | \(=\) | \(\ds 379\) | ||||||||||||
| \(\ds 3^2 + 5^2 + 5^2\) | \(=\) | \(\ds 59\) | ||||||||||||
| \(\ds 3^2 + 5^2 + 7^2\) | \(=\) | \(\ds 93\) | ||||||||||||
| \(\ds 3^2 + 7^2 + 7^2\) | \(=\) | \(\ds 107\) | ||||||||||||
| \(\ds 3^2 + 7^2 + 11^2\) | \(=\) | \(\ds 179\) | ||||||||||||
| \(\ds 3^2 + 7^2 + 13^2\) | \(=\) | \(\ds 227\) | ||||||||||||
| \(\ds 3^2 + 7^2 + 17^2\) | \(=\) | \(\ds 347\) | ||||||||||||
| \(\ds 3^2 + 7^2 + 19^2\) | \(=\) | \(\ds 419\) | ||||||||||||
| \(\ds 3^2 + 11^2 + 11^2\) | \(=\) | \(\ds 251\) | ||||||||||||
| \(\ds 3^2 + 11^2 + 17^2\) | \(=\) | \(\ds 467\) | ||||||||||||
| \(\ds 3^2 + 11^2 + 19^2\) | \(=\) | \(\ds 491\) | ||||||||||||
| \(\ds 3^2 + 13^2 + 13^2\) | \(=\) | \(\ds 347\) | ||||||||||||
| \(\ds 3^2 + 13^2 + 17^2\) | \(=\) | \(\ds 467\) | ||||||||||||
| \(\ds 3^2 + 17^2 + 17^2\) | \(=\) | \(\ds 587\) | ||||||||||||
| \(\ds 3^2 + 17^2 + 19^2\) | \(=\) | \(\ds 659\) |
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:
| \(\ds a^2\) | \(\equiv\) | \(\ds 1\) | \(\ds \pmod 3\) | |||||||||||
| \(\ds b^2\) | \(\equiv\) | \(\ds 1\) | \(\ds \pmod 3\) | |||||||||||
| \(\ds c^2\) | \(\equiv\) | \(\ds 1\) | \(\ds \pmod 3\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a^2 + b^2 + c^2\) | \(\equiv\) | \(\ds 1 + 1 + 1\) | \(\ds \pmod 3\) | Modulo Addition is Well-Defined | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds a^2 + b^2 + c^2\) | \(\equiv\) | \(\ds 0\) | \(\ds \pmod 3\) | Definition of Congruence Modulo Integer | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds a^2 + b^2 + c^2\) | \(\equiv\) | \(\ds 0\) | \(\ds \pmod 3\) | Definition of Congruence Modulo Integer |
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.
$\blacksquare$