Does $a^2 + b^2 = 2 c^2$ have any integer solution?

Question

Does the equation $a^2 + b^2 = 2 c^2$ have any integer solution with $|a| \neq |b|$?

I think not, because of these equations for pythagorean triplets: $$\left(\frac{a}{c}\right)^2+\left(\frac{b}{c}\right)^2=1$$ $$x=\frac{1-t^2}{1+t^2}, y=\frac{2t}{1+t^2}$$

I think I would need to multiply $x$ and $y$ by $\sqrt{2}$ and they would never be rational numbers.

Answer 1

It is equivalent to solving the circle equation $x^2 + y^2 = 2$ in rational numbers.

Take one point $(1,1)$ on this curve and consider the line with slope $t$ passing through that line: $y = t x - t + 1$, substitute this in $x^2 + (t x - t + 1)^2 - 2 = 0$ and divide out $(x-1)$ to find the second point of intersection between this line and the circle, $(t^2 + 1)x + (-t^2 + 2t + 1) = 0$.

We have parametrized all nontrivial solutions $$x = a/c = \frac{-t^2 + 2t + 1}{t^2 + 1}$$

For example $t = 22/7$ gives us $a = 127$, $c = 533$ and $127^2 + 743^2 = 2\cdot 533^2$.

Answer 2

There are infinitely many solutions with $a=1$. We want to solve $b^2-2c^2=-1$. Armed with the solution $b=c=1$ and the fact that $3^2-2\cdot 2^2=1$ we can use the Brahmagupta-Fibonacci identity to find more. Given a solution $(i,j)$ the next one is $(3i+4j,2i+3j)$, leading to the sequence $(7,5), (41,29),(239,169),(1393,985)\ldots$ You can negate either or both members of these pairs because $b$ and $c$ are squared.

Answer 3

Particular case where $b=1$ and $a^2+1=2c^2$ is related to Euler's equation $ 2y^4-1=z^2$.Euler found $(z, y)=(1, 1)$ and $(239, 13)$. Lungern proved these are the only solutions. But $2c^2-1=a^2$ has infinite solutions and I found a recursive relation:

$c=1, 5, 169, 5741, . . . , (34 c_{i-1}-c_{i-2})$

$a=1, 7, 239, 8119, . . . , (34 a_{i-1}-a_{i-2})$

Other set of theses numbers are:

$c=1,29 ,985, 33461 . . , (34 c_{i-1}-c_{i-2})$

$a=1, 41, 1393, 47321 . . . ,(34 a_{i-1}-a_{i-2})$

Where $i$ is the index of the number .The algorithm is based on assumption that: $c-a=d_1$ and $c-b=d_2$ and general form $a^2+b^2=c^2$ which led to find $119^2+120^2=169^2=13^4$. This was my attempt to fond a solution for Euler's equation. The set of numbers showed was the result of this algorithm.

Answer 4

$a^2+b^2=2c^2\tag{1}$
We can get infinitely many integer solutions by Diophantus's method below.
Substitute $a=t+p, b=t+q, c=t+r$ to equation $(1)$, then we get $$(2p+2q-4r)t+p^2-2r^2+q^2=0$$ Let $t = -1/2(p^2-2r^2+q^2)/(p+q-2r)$.
Then we get a parametric solution below.
$a = p^2+(-4r+2q)p+2r^2-q^2$
$b = p^2-2r^2-q^2-2pq+4qr$
$c = p^2+2r^2+q^2-2pr-2qr$
p,q,r are arbitrary.
Thus, we get infinitely many integer solutions.

Example:

                     [p,q,r]    [a,b,c]  

                     [3, 2, 1], [7, 1, 5]
                     [4, 3, 1], [17, 7, 13]
                     [5, 2, 1], [23, 7, 17]
                     [5, 4, 1], [31, 17, 25]
                     [6, 3, 1], [41, 1, 29]
                     [6, 5, 1], [49, 31, 41]
                     [7, 2, 1], [47, 23, 37]
                     [7, 6, 1], [71, 49, 61]

Answer 5

Suppose $(x,y,z)$ is any Pythagorean triple. Then: $$(x-y)^2+(x+y)^2=2x^2+2y^2=2z^2$$