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I have found lots of numbers that can be done two ways but can't seem to find any that are done three or four ways. I believe there would be a way to use complex numbers to help with this.

Bill Dubuque
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Danny Pilcher
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  • What does it mean for a number to be done? Some examples of what you have found, to show what you mean, would help to clarify your post. – Lee Mosher Nov 13 '23 at 12:12
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    The place to start is $$a^2 + 2b^2 = r^2 + 2s^2 \iff [(a-r)(a+r) = 2[(s-b)(s+b)].$$ – user2661923 Nov 13 '23 at 12:12
  • Take a look at the (general) Brahmagupta identity, equations (3) and (4) on that page. – Jaap Scherphuis Nov 13 '23 at 12:13
  • @LeeMosher OP intends that $33$ can be done by either $5^2 + 2(2)^2~$ or $~1^2 + 2(4)^2.~$ – user2661923 Nov 13 '23 at 12:14
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    I would like to believe you, but I would rather that the OP be clearer. This is a situation where the burden of clarity is on the OP. @user2661923 – Lee Mosher Nov 13 '23 at 12:16
  • @LeeMosher I used to embrace the perspective that you are indicating. However, I got tired of (well over half the time) getting no response from the original poster. – user2661923 Nov 13 '23 at 12:18
  • That's an easy situation to deal with: vote to close. – Lee Mosher Nov 13 '23 at 12:19
  • I am trying to find numbers that can be made but using this formula with different integers such as 33. i am interested in finding numbers that can be created with 3 or 4 different ways for different values of a and b. I hope this is clearer. – Danny Pilcher Nov 13 '23 at 12:40
  • @user2661923 thank you i will look into this further. – Danny Pilcher Nov 13 '23 at 12:50
  • $33$ can be done in two ways because $33=3\cdot 11$, and $3=1^2+2\cdot 1^2$ and $11=3^2+2\cdot 1^2$ and Brahmaguptra gives two ways to combine them. Multiply more such factors to get a product that has more ways to write it in that form. – Jaap Scherphuis Nov 13 '23 at 12:54
  • I wrote the number of integer solutions to $x^2+ay^2=n$ for $a\in 1,2,3,4$ over here: https://math.stackexchange.com/questions/2834085/how-many-integer-pairs-satisfy-the-ellipse-x2ay2-r – Mason Nov 13 '23 at 21:32

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COMMENT.-I think what you want is to find more of two integer solutions $(h,k)$ to the equation $$x^2+2y^2=n\qquad(1)$$ where $n\in\mathbb N$. Since you do have for all $n$ the equation of an ellipse it is enough to look at $x$ and $y$ non-negative and the other ones are the corresponding symmetric.

For example, if $$x^2+2y^2=81$$ you have the three "main" solutions $(a,b)=(3,6),(7,4),(9,0)$ from which you do have $10$ integer solutions because if $(a,b)$ is solution so is $(\pm a,\pm b)$.

In order to have more of three "main" solutions you have to consider greather values of $n$ in the equation $(1)$ above.

Piquito
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