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$a$, $b$ and $c$ of all Pythagorean triplets can be written in the form $$ \begin{split} a &= 2mn\\ b &= m^2-n^2 \\ c &= m^2+n^2 \end{split} $$ where $m$ and $n$ are natural numbers. For any natural number $m$ and $n$, this set of equations will give a Pythagorean triplet. And all Pythagorean triplets satisfy this set of equations.

Can $a$, $b$ and $c$ of all triplets satisfying the equation $$a^2+b^2=2c^2$$ where $a$, $b$ and $c$ are natural numbers, be written as a set of equations as for the Pythagorean triplets?

So, I need a set of equations that generates triplets that satisfy the equation $a^2+b^2=2c^2$ for any natural numbers I plug into the set of equations. Also, every natural number triplets satisfying the equation $a^2+b^2=2c^2$ must satisfy the set of equations.

I tried to derive the set of equations myself, no attempts have been successful yet.

I would like to have the proof of the set of equations, (otherwise I won't know if every triple will satisfy the set of equations)

Any comments that helps to give an insight into solving the problem are really appreciated.

  • Have you tried modifying the proof of the derivation of the solution for the standard Pythagorean triples? If so, what did that yield? Could you show your attempts? – Clayton Jan 24 '23 at 15:26
  • @ClaytonYes, I have tried that, but I got some equations with sqrt2 and and things like that in it. And it didn't guaranteed that a, b and c are natural numbers. –  Jan 24 '23 at 15:33
  • Sávio has already given you a complete answer. – Clayton Jan 24 '23 at 15:39
  • https://math.stackexchange.com/questions/1767109/x2y2-2z2-positive-integer-solutions/1767719#1767719 – individ Jan 24 '23 at 15:48
  • @individ Oh! I was unaware of that. Anyway, I got the exact same answer as one of the answer given to that question with the help of Sávio's reply. –  Jan 24 '23 at 16:08

3 Answers3

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Hint: For $a^2+b^2=2c^2$, observe that $a, b$ have the same parity. Therefore there exist integers $u, v$ such that $a = u+v$ and $b = u-v$. Expand...

Sávio
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  • When I did what you said, I got c=sqrt(u^2 +v^2), which isn't guaranteed to be a natural number. I need a, b and c to be a natural number. –  Jan 24 '23 at 15:19
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    So you need $(u, v, c)$ to be a Pythagorean triple. – Sávio Jan 24 '23 at 15:22
  • @Hisham they can be natural, simply because $a \equiv b \mod 2$ – Dstarred Jan 24 '23 at 15:26
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    I got a=u+v, b=u-v and c=sqrt(u^2+v^2). Since, u, v and c are Pythagorean triplets, substituting 2mn for u, m^2-n^2 for v and m^2+n^2 for c and some cancelling later, I got to the final answer: a=m^2-n^2+2mn, b=n^2-m^2+2mn and c=m^2+n^2 –  Jan 24 '23 at 16:15
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Here is a proof on why natural solutions exist for $a, b, c$, as a matter of fact, infinite solutions.

$$a^2 + b^2 \equiv 0 \mod 2 \to (a + b)^2 - 2ab \equiv 0 \mod 2 \\ \quad \\ \implies a + b \equiv 0 \mod 2 $$

More importantly, this suggests $$a \equiv b \mod 2 \implies \text{ both $a - b$ and $a + b$ must be even}$$

Then there are such integers $x, y$ which satisfy $a - b = 2x$ and $a + b = 2y$.

Hence,

$$4x^2 + 4y^2 = (a + b)^2 + (a-b)^2 = 2(2c^2) \\ \quad \\ \implies x^2 + y^2 = c^2 * \qquad \square$$

For better understanding, $a, b$ have to be both even or odd, hence have the same remainder when divided by $2$. Furthermore, $*$ suggests there are infinite solutions to $a^2 + b^2 = c^2$ as there are infinite integer pythagorean triples.

Dstarred
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    I knew there are infinite natural number solutions to it. But what I wanted was algebraic form for all natural numbers a, b and c which satisfies the equation. And I have got it. I liked the proof though. –  Jan 24 '23 at 15:56
  • @Hisham No problem – Dstarred Jan 24 '23 at 16:01
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Noting that $a^2+b^2=2c^2 \Rightarrow a $ and $b$ are of same parity. Hence there exists natural numbers $u$ and $v$ such that $$a+b=2u \textrm{ and } a-b=2v$$ Then $$a=u+v \textrm{ and }b=u-v$$ $$ \begin{gathered} (u+v)^2+(u-v)^2=2 c^2 \Rightarrow u^2+v^2=c^2 \end{gathered} $$ There exists Pythagorean triple such that $$ \begin{aligned} & \left\{\begin{array}{l} u=2k m n \\ v=k(m^2-n^2) \\ c=k(m^2+n^2) \end{array}\right. \Rightarrow \left\{\begin{array}{l} a=k(2 m n+m^2-n^2 )\\ b=k(2 m n-m^2+n^2) \\ c=k(m^2+n^2) \end{array}\right. ,\\ & \end{aligned} $$ where $k,m,n \in N$.

Lai
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