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I manage to find a method to compute integer solutions of the equation: $$ \frac{x^2}{a^2}+\frac{y^2}{b^2} = k $$ where $$(a,b)\in\mathbf{Z}\times\mathbf{Z}$$ and $$k\in\mathbf{N}$$ I would like to know if there is already a known solution to this problem or if I just proved a new method to compute those solutions.

My approach:

first consider the equation:

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2} = k $$ and multiply left and right hand side by $\gcd(a^2,b^2) = (\gcd(a,b))^2=d$. We will obtein an equivalent equation of the form: $$ \frac{x^2}{a_1^2}+\frac{y^2}{b_1^2} = k_1 $$ where $k_1 = k\cdot d$ and $\gcd(a_1,b_1) = 1$. Now we will consider the equations: $$ \frac{x^2}{a_1^2}+\frac{y^2}{b_1^2} = 1 $$ and: $$ x^2+y^2 = k_1 $$ The first one has only 4 integer solutions since $a_1,b_1$ are coprime. The solutions are: $(\pm a_1,0),(0,\pm b_1)$. Now consider a generic solution of the second equation, call it: $(x_n,y_n)$. Then we have the following result: $$ \Bigl(\langle (x_n,y_n) , (\pm a_1,0) \rangle , \langle (x_n,y_n), (0,\pm b_1) \rangle \Bigr) $$ is a solution of the main equation. Or simply: $(\pm a_1x_n, \pm b_1y_n)$ is a solution. (I used scalar product notation because I'm working to a more general case where it plays a fundamental role). For example, consider the equation: $$ \frac{x^2}{9}+\frac{y^2}{36} = 45 $$ rewrite it as: $$ \frac{x^2}{1}+\frac{y^2}{4} = 405 $$ The integer solutions of $\frac{x^2}{1}+\frac{y^2}{4} = 1$ are $(\pm 1,0),(0,\pm 2)$, while the integer solutions of $x^2+y^2 = 405$ are of the form: $(\pm18,\pm9),(\pm9,\pm18)$. Then the solutions of the main equation are of the form: $$ (\pm18,\pm18) \mbox{ or } (\pm9,\pm36) $$

MathRevenge
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    See if this is of help: https://mathworld.wolfram.com/DiophantineEquation2ndPowers.html – Intelligenti pauca Jul 20 '22 at 07:05
  • It would be helpful to post your method so people can tell you if it's a new method. If you post your method, it will be yours until the end of time. – user25406 Jul 20 '22 at 12:34
  • I added my method with an example, let me know and thank you in advance. – MathRevenge Jul 20 '22 at 15:17
  • There are solutions if you can imagine - $k$ - as the sum of two squares. https://math.stackexchange.com/questions/1767109/x2y2-2z2-positive-integer-solutions/1767719#1767719 – individ Jul 20 '22 at 15:50
  • Yes indeed! Also, if $k = a^2+b^2$ for some integers a,b, we have that the number of solutions of $\frac{x^2}{a^2}+\frac{y^2}{b^2} = k$ is the same as $x^2+y^2 = k$. – MathRevenge Jul 21 '22 at 16:30

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