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As the title states, the question tasks me with finding all the integral solutions of the equation under the specified constraints. I have an idea of where to start due to a somewhat similar problem in my notes, but I'm having trouble adapting it to this new equation, and knowing where I need to change my process. My professor has provided the hint: "Use the circle $^2 + ^2 = 2$ and the lines passing through the point $(1,1)$." Here's what I have so far:

$x^2 + y^2 = 2z^2$

$\frac{x^2}{z^2} + \frac{y^2}{z^2} = 2$

Let $X = \frac{x}{z}$ and $Y = \frac{y}{z}$

$X^2 + Y^2 = 2$

So we now have a circle with an origin at $(0,0)$ and radius of $\sqrt{2}$. I then drew up the circle and the line passing through $(1,1)$.

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So the slope $\lambda$ of this line would be: $\lambda = \frac{Y-1}{X-1}$. This is the part where I get lost, in class we went off on a tangent related to these problems and I'm having trouble knowing exactly how to proceed. I have an idea of what the final answer will look like. The form we found for $x^2 + y^2 = z^2$ was: $(x,y,z) = (a^2 - b^2, 2ab, a^2 + b^2)$ Any help would be greatly appreciated!

Drew Pesall
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3 Answers3

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start with a primitive Pythagorean triple $x^2 + y^2 = z^2.$ there is a recipe for these. Then $$ (x+y)^2 + |x-y|^2 = 2 z^2 $$ And those are all

Will Jagy
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  • The OP asked for how to continue his solution, not another way to solve the problem. But your way is a good one ;) – Trebor Apr 17 '19 at 23:55
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Since you have worked out the slope, work out the equation of the $x$-coordinates of the two intersections of the line and the circle, then since you know the coord of one, use Viete's theorem.

Trebor
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The trick to this one is to create three new variables $$ a = x+y\\ b = x-y\\ c = 2z $$ Then $x^2+y^2 = 2z^2 \implies a^2+b^2 = c^2$

Temporarily ignoring the condition that $\gcd(x,y,z) = 1$ we can now list all solutions of $a^2+b^2 = c^2$ as $$ a = k(n^2 - m^2)\\ b = 2kmn\\ c = k(n^2 + m^2) $$ with $m \neq n \pmod 2$ and $\gcd(m,n) = 1$.

To get back to $x,y,z$ we have $$ x = \frac{b+a}2 = k\frac{(n^2 + 2mn - m^2)}2 \\ y = \frac{b-a}2 = k\frac{(m^2 + 2mn - n^2)}2 \\ z = \frac c2 = k\frac{(n^2 + m^2)}2 $$ However, since $m$ and $n$ are of opposite parity, $(n^2 + m^2)$ is odd, so in order for $z$ to be an integer, $k$ must be even: $k = 2p$.

Then the generic triplet is $$ x = p(n^2 + 2mn - m^2) \\ y = p(m^2 + 2mn - n^2) \\ z = p(n^2 + m^2) $$ $x$ and $y$ are not coprime unless $p = 1$. So we have $$ x = n^2 + 2mn - m^2 \\ y =| m^2 + 2mn - n^2| \\ z = n^2 + m^2 $$ Finally, we can ask when $\gcd(x,y) = 1$ but it is easier to ask when $(x,z) = 1$ which is equivalent for our purposes.

The difference $z-x = m(m-2)$ is a multiple of $m$ and thus is coprime with $z$ because $\gcd(n^2,m) = 1)$ So all such triplets are coprime, and that is our answer:

$$ x = n^2 + 2mn - m^2 \\ y =| m^2 + 2mn - n^2| \\ z = n^2 + m^2 $$ with $ n,m\in \Bbb N \wedge n > m \wedge m \neq n \pmod 2 \wedge \gcd(m,n) = 1$.

Mark Fischler
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