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a long time ago I posted a question to find a general solution to a modified Pythagorean equation, mainly $a^2+b^2=2c^2$ that question was eventually answered. But now I need more help.

I now have 3 separate equations, and I would like a general solution for each (independently for each single version).

$$2a^2+b^2-c^2=2d^2 \tag {eq.1}$$ $$2a^2+b^2+c^2=4d^2 \tag {eq.2}$$ $$3a^2+2b^2-c^2=4d^2 \tag {eq.3}$$

Any help would be appreciated since I have no idea how I would come up with general solutions to these. Thank you in advance.

Also, I will only consider any solution where all numbers $a,b,c,d$ in all equations are pairwise different (and all are not zero).

PS. for those asking about the tags, yes this is related to 3x3 magic squares.

spydragon
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  • $a = d, b = c$ (but with $a\ne b$) is another set of trivial solutions to the first equation. – Paul Sinclair Jan 30 '23 at 18:19
  • @PaulSinclair I really should have noticed that. Thanks, I will modify the question to be more accurate. – spydragon Jan 31 '23 at 20:14
  • Okay - FYI, "Diophantine equations" like these are tricky. The techniques that solve one equation are often not applicable to the next. There isn't some established method for solving them. I would guess that the first two should not be too hard to solve because of certain symmetries, but the third might be. – Paul Sinclair Jan 31 '23 at 20:56
  • It reminds me of some steps when I was anylyzing the MSqofSq-s problem, but don't have an exact relation at hand. So I can't help at the moment. Anyway, perhaps this answer here in MSE is informative for you: https://math.stackexchange.com/a/175762/1714 I had as well put my analyses in a small essay, but where I still could not solve the system of coupled (generalized) Pell- and Pythagorean-like equations... See https://go.helms-net.de/math/divers/Helms_2022_magicSquareOfSquares2_draft.pdf Nice subject- sorry cannot help more... – Gottfried Helms Feb 03 '23 at 12:16
  • User @individ has a long web-page (at art-of-problem-solving) on sums of sqaures, pythagorean equations generalized etc. Perhaps you can find something there. Start at https://artofproblemsolving.com/community/c3046h1471815 and see what is there. – Gottfried Helms Feb 03 '23 at 13:58
  • There has been a couple of related questions at around nov 2022 here in MSE, perhaps this one could be a starting point for further investigations: https://math.stackexchange.com/questions/4546172/ – Gottfried Helms Feb 03 '23 at 22:04

1 Answers1

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The first equation can be rearranged to $$2a^2-2d^2 = c^2 - b^2\\2(a-d)(a+d) = (c-b)(c+b)$$

If $d > a$, then we can exchange the values of $a$ and $d$, and exchange the values of $c$ and $b$ to get a solution with $a > d$. So it is sufficient to assume $a > d$. This makes the left side positive, so the right side must be positive as well. Thus $c > b$.

Now at least one of $c-b$ and $c+b$ must be even. But they differ by $2b$, so both are even. Let $r = \frac{c+b}2, s= \frac{c-b}2$. Then $$(a-d)(a+d) = 2rs$$ So one of $a-d$ and $a+d$ must be even, and thus both. Let $p = \frac{a+d}2, q = \frac{a-d}2$. Now we have $$2pq = rs$$

Solutions to this equation are obvious. Pick arbitrary $p, q$ with $0 < q < p$. Then pick $r > s$ to be any factorization of $2pq$ (including $r = 2pq, s= 1$).

Then set $$a = p+ q\\b=r-s\\c = r+s\\d=p-q$$ to get a solution to $2a^2 + b^2 - c^2 = 2d^2$. Every non-trivial solution to the equation with $a > d$ can be obtained in this way, and every solution with $a < d$ can be obtained by exchanging the values of $(a,c)$ with those of $(d, b)$. The trivial solutions can be obtained by allowing $p=q, r=s$ or $q = s = 0$ as well.

Paul Sinclair
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