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

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The discussion centers on the impossibility of finding integers a, b, and c such that a^2 + b^2 equals 3 times a perfect square (3*c^2). This conclusion is derived from the sum of two squares theorem, which states that if an integer's prime decomposition includes 3 raised to an odd power, it cannot be expressed as the sum of two squares. Since 3c^2 contains an odd power of 3, it follows that it cannot be represented as a sum of two squares. The participants express uncertainty about proving this result without relying on the sum of two squares theorem, highlighting its complexity in number theory. Ultimately, the theorem provides a definitive answer to the posed question.
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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.
 

Thread 'About the definition of topological manifold using closed sets'

We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.
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