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

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SUMMARY

The discussion confirms that there are no integers a, b, and c such that the equation a² + b² = 3c² holds true. This conclusion is derived from the sum of two squares theorem, which states that if the prime factorization of an integer includes 3 raised to an odd power, that integer cannot be expressed as the sum of two squares. Since 3c² contains 3 raised to an odd power, it follows that 3c² cannot be represented as the sum of two squares.

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

  • a^2 + b^2 = 3*c^2
 
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That depends on what [math]a, b[/math] and [math]c[/math] are.
 
MathLover_James said:
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.
 

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