Pythagorean Investigation

 

Here’s a challenge @jamestanton tweeted yesterday:

Visually, Tanton is asking if there is an integer right triangle (like the standard version shown on the left below) for which the integer triangle on the right exists.

Pythag

The algebraic equivalent to this question is, for some a^2+b^2=c^2, does there exist a Natural number d so that b^2+c^2=d^2?

I invoked Euclid’s formula in my investigation to show that there is no value of d to make this possible.  I’d love to hear of any other proof variations.

INVOKING EUCLID’S FORMULA

For any coprime natural numbers m & n where m>n and m-n is odd, then every primitive Pythagorean triple can be generated by \left\{ m^2 - n^2, 2mn, m^2 + n^2 \right\}.

For any Natural number kevery Pythagorean triple can be generated by \left\{ k \cdot \left( m^2 - n^2 \right), k \cdot \left( 2mn \right), k \cdot \left( m^2 + n^2 \right) \right\}.

The generator term k \cdot \left( m^2 + n^2 \right) must be the original hypotenuse (side c), but either k \cdot \left( m^2 - n^2 \right) or k \cdot \left( 2mn \right) can be side b.  So, if Tanton’s scenario is true, I needed to check two possible cases.  Does there exist a Natural number d such that

d^2 = \left( k \cdot \left( m^2 - n^2 \right) \right)^2 + \left( k \cdot \left( m^2 + n^2 \right) \right)^2 = 2k^2 \left( m^4 + n^4 \right)

or

d^2 = \left( k \cdot \left( 2mn \right) \right)^2 + \left( k \cdot \left( m^2 + n^2 \right) \right)^2 = k^2 \left( m^4 +6m^2n^2 + n^4 \right)

is true?

EVALUATING THE POSSIBILITIES

For the first equation, there is a single factor of 2 on the right, and there is no way to extract an odd number of factors of 2 from \left( m^4 + n^4 \right) or k^2, so 2k^2 \left( m^4 + n^4 \right) can’t represent a perfect square.

For the second equation, there is no way to factor \left( m^4 +6m^2n^2 + n^4 \right) over Integers, so k^2 \left( m^4 +6m^2n^2 + n^4 \right) can’t be a perfect square either.

Since neither equation can create a perfect square, there is no Natural value of d that makes {b, c, d} a Pythagorean triple.  Tanton’s challenge is impossible.

Does anyone have a different approach?

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