All natural number solutions for the equation $a^2+b^2=2c^2$

Question

$a$, $b$ 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.

Answer 1

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...

Answer 2

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,

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

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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.

Answer 3

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 $k,m,n \in N$.

Answer 4

Let's look at the general case.

$$aX^2+bY^2=cZ^2$$

When the coefficients of this equation are not represented in the form of squares. Then a funny pattern emerges.

If we know some solution to the equation and we know any solution to the Pell equation. Then we can always construct the following solution to this equation.

$$\left\{\!\begin{aligned} & ax^2+by^2=cz^2 \\ & cp^2-as^2=1 \end{aligned}\right. $$

$$f=\sqrt{ac(sz+px)^2+by^2}$$

$$\left\{\!\begin{aligned} & X=cxp^2+fs+czps \\ & Z=axps+pf+azs^2 \end{aligned}\right. $$

Formally, the relationship between the solvability of the Lagrange equation in general and the solvability of the equivalent Pell equation is obtained. Numbers can have any signs.