Can the Sum of Two Squares Ever Equal 3 Times a Perfect Square?

[MathLover_James]

There can't be a, b and c integers such that:

• a^2 + b^2 = 3*c^2

 

[Greg]

That depends on what \(\displaystyle a, b\) and \(\displaystyle c\) are.

 

[Opalg]

There can't be a, b and c integers such that:

• a^2 + b^2 = 3*c^2

This is a consequence of the sum of two squares theorem. This says (among other things) that if the prime decomposition of an integer $n$ contains $3$ raised to an odd power then $n$ cannot be the sum of two squares. Since the number $n = 3c^2$ has an odd power of $3$ in its prime decomposition, the theorem says that it cannot be the sum of two squares.

I don't know whether you can prove the result about $3c^2$ without using the sum of two squares theorem, which is a fairly deep result in number theory.