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I have come across the OEIS series A031363. It has this description:

Positive numbers of the form $x^2+xy-y^2$; or, of the form $5x^2-y^2$.

So it is saying that all numbers that can be written in the form $x^2+xy-y^2$ can be written as $5x^2-y^2$.

I am a beginner at Number Theory, so any tips or hints on how to proof it would be great.

EDIT: I have done a tiny computer search on it, and confirmed that all numbers (< 10000) that can be written in the first form can also be written in the second form. So it is saying that both form is identical.

Gareth Ma
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  • Or is it just saying that the sequence is the union of the sets of the numbers of these forms? – Angina Seng Feb 26 '18 at 06:41
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    @LordSharktheUnknown If so, the semicolon is very strange. – Arthur Feb 26 '18 at 06:43
  • Have you looked through the links provided on the site? – SK19 Feb 26 '18 at 06:45
  • @SK19 What do you mean? – Gareth Ma Feb 26 '18 at 06:47
  • @LordSharktheUnknown Both forms are identical. Please read the edit. – Gareth Ma Feb 26 '18 at 06:47
  • On the OEIS site, there is a category called "Links". There you can look up research related to these sequences. On another note: Proving that number $n$ up to some specific $N$ that is representable as $x^2+xy-y^2$ is also representable as $5x^2-y^2$ doesn't show the converse and also doesn't show that the forms are identical for all $N$. – SK19 Feb 26 '18 at 06:57
  • I think this is actually a pretty good question! The two expressions of course give different answers for a fixed pair $(x,y)$, but it seems like for every pair there is another pair that solves the other equation! For example, if we plug in $x=y=2$ in the first expression, the second gets the same value if $x= -233 $ and $y= -521 $. Tricky ... – Matti P. Feb 26 '18 at 06:58
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    Initial idea: If $x$ is even, that is $x=2a$, then $$x^2 + xy - y^2 = 4a^2 + 2ay - y^2 = 5a^2 - (a-y)^2$$. EDIT: upon further thought this says that every number of the latter form is necessarily of the former form, but not necessarily the reverse yet. – Bob Krueger Feb 26 '18 at 06:59
  • Also, a good strategy for generating some sort of pattern: you've already generated a lot of test cases, and presumably have pairs $(x,y)$ for the first form that correspond to pairs $(x,y)$ for the second form. Why not plot these values and see if there is a simple (say, linear, as my previous comment suggests) relationship? – Bob Krueger Feb 26 '18 at 07:05
  • Sure. @BobKrueger – Gareth Ma Feb 26 '18 at 07:16

2 Answers2

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If one of $x$, $y$ is even, WLOG let $x = 2a$. Same as what Bob Krueger said, $x^2 + xy - y^2 = 5a^2 - (a-y)^2$

If both $x$ and $y$ are odd, then $x + y$ and $x + 3y$ are even. We can check $x^2 + xy - y^2 = 5\left(\frac{x + y}{2}\right)^2 - \left(\frac{x + 3y}{2}\right)^2$

Guest
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    Inverting the last transformation, we find that $$5u^2-v^2=(3u-v)^2+(3u-v)(v-u)-(v-u)^2$$ so that the converse holds too. – saulspatz Feb 26 '18 at 07:40
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    I'm afraid I don't see how this works when $y$ is even. If $y=2a,$ I get $$x^2+2ax-4a^2=2x^2-(2a-x)^2,$$ and I don't see how to put in the the required form. What am I missing? – saulspatz Feb 26 '18 at 15:27
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It seems to me that there is a case missing in the accepted answer. I don't see how we can take $x=2a$ "without loss of generality," because the form $x^2+xy-y^2$ is not symmetric in $x$ and $y.$ Indeed, when $y=2a,$ we have $x^2+xy-y^2 = x^2+2ax-4a^2=2x^2-(2a-x)^2,$ and I don't see how to put this in the required form.

However, when $y$ is even, we have $$5\left(x-\frac{y}{2}\right)^2-\left(2x-\frac{3y}{2}\right)^2=x^2+xy-y^2,$$

so the theorem is true is this case too.

As I noted in a comment on the accepted answer, the identity $$5u^2-v^2=(3u-v)^2+(3u-v)(v-u)-(v-u)^2$$ establishes the converse.

saulspatz
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