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I don't know any identity that'd help me simplify it. I know of Brahmagupta's identity and tried using but no good. Any hints? Edit: So far I've tried various things, $(x^2+1/2)^2 + (y^2 +1/2)^2 -1/2$ This doesn't get me anywhere. I tried solving in x as a quadratic so as to get an insight or something but that didn't help too. I even tried writing general polynomials of degree too and then getting the solution by comparing coefficients, but that wasn't elegant and was too computational.

Rodrigo de Azevedo
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Sajid Rizvi
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    Why all the upvotes? It's a cool problem, sure, but it's not a very good post. – Arthur May 12 '18 at 14:31
  • Those are two different reasons for upvoting :) – Hans Engler May 12 '18 at 14:32
  • @HansEngler That's what favourite is for, in my opinion. – Arthur May 12 '18 at 14:35
  • @Arthur: Because there are followers of this portal that have a low level in mathematics. Some of them look for difficult problems only to see if they are answered, while others raise any insignificant little thing. In addition, while some give the approval of the best answer to the first one received without waiting for others to compare, others do not place any recognition to the resolution of the difficult problem (which have demanded of efforts) that they have proposed. – Ataulfo May 12 '18 at 14:43
  • Is something wrong in my post? Do I need to edit something? – Sajid Rizvi May 12 '18 at 15:04
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    @SajidRizvi The post needs to contain details of what you've tried so far, namely your efforts. – Ѕᴀᴀᴅ May 12 '18 at 15:09
  • Nothing at all. The problem is not with you, dear friend. Personally I think your problem is not without a certain elemental interest. – Ataulfo May 12 '18 at 15:09
  • The most common downvote reason for question posts on this site is, as far as I can tell, that they do not provide context. Questions on the level of high school homework exercises, where the post contains only the problem statement (and possibly "how do I solve this" or "any help appreciated" or similar) attract loads of downvotes. The biggest difference between those posts and yours is the difficulty of the problem. I'm mostly venting my annoyance at the voting inconsistency on this site. – Arthur May 12 '18 at 15:16
  • @AlexFrancisco I cannot follow this argument. The author mentioned what he/she tried, perhaps it would be better to write this effort out, however. – Peter May 12 '18 at 15:27
  • I have edited my post. If you all can suggest identities that may help, I'll be glad. Also I'm not a frequent stackexchange user, so I don't understand the upvote down vote thing very much. Apologies for any inconveniences. – Sajid Rizvi May 12 '18 at 15:33
  • @SajidRizvi So close, yet so far. – Peter May 12 '18 at 15:39
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    This is maybe useful for you $$(2ab)^2+(2bc)^2+(c^2-a^2-b^2)^2=(a^2+b^2+c^2)^2$$. Try first and if you don't solve your problem, I'll try to do it. – Ataulfo May 12 '18 at 15:43
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    Are you sure that the problem has a solution in $\mathbb Q[x,y]$? Where does the problem come from? – Ataulfo May 12 '18 at 16:13

4 Answers4

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let real constant $$ t = \sqrt{\sqrt 8 - 2} $$ so that $$ \frac{t^4}{4} + t^2 = 1. $$ Then $$ \left(-x^2 + \frac{t^2}{2} y^2 \right)^2 + \left( t y^2 + x \right)^2 + \left( txy - y \right)^2 = x^4 + y^4 + x^2 + y^2 $$

ADDED: apparently this was asked and answered six years ago, and I commented there.

Will Jagy
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As an application of Legendre's three squares theorem, this problem has no solution in $\mathbb Z[x,y]$. In fact, suppose $$f(x,y)=x^4+y^4+x^2+y^2=(h_1(x,y))^2+(h_2(x,y))^2+(h_3(x,y))^2$$ so one has $$f(1,3)=92=(h_1(1,3))^2+(h_2(1,3))^2+(h_3(1,3))^2$$ But $$92=4(2\cdot8+7)$$

This is impossible; for the above theorem, $92$ can not be representable as a sum of three squares (that one or two of the $h_i(1,3)$ be zero is easily discarded).

Ataulfo
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  • Well, fortunately OP didn't say the coefficients had to be integers (or even rational). – mjqxxxx May 12 '18 at 16:49
  • What the OP probably thought is $\sum a_i P_i^2$ where $a_i>0$ (rationals) and $P_i$ polynomials (probably rational coefficients). – orangeskid May 12 '18 at 16:57
  • Do not you think it should be forbidden in Stack Exchange to put downvotes without explaining why? – Ataulfo May 12 '18 at 17:20
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More of a comment: Hilbert proved that this was possible in 1888 (a positive definite quartic in two variables can always be written as a sum of squares of three polynomials). For instance, see the discussion of the proof in [1]. But it is a highly nontrivial result, and the proof is not especially constructive.

[1] W. Rudin. Sums of squares of polynomials. Amer. Math. Monthly, $107(9):813–821$, $2000$.

mjqxxxx
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  • Can you put your reference as a link, please? It is impossible that Hilbert was wrong and that his error has not been detected in all the time elapsed so that, in effect, he must have referred to polynomials with real coefficients. In any case, I believe that the O.P. implicitly referred to rational coefficients. He would have the last word on this point. Regards. – Ataulfo May 12 '18 at 17:03
  • The Hilbert result is for polynomials with real coefficients. – mjqxxxx May 12 '18 at 17:42
  • got it....................... – Will Jagy May 12 '18 at 18:27
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Just an observation, one considers a writing $$f(x,y)= (x,y,x^2, xy, y^2) M (x,y,x^2, xy, y^2)^t$$ where $M$ is a symmetric $5\times 5$ matrix. Assume $M$ is positive semidefinite. Diagonalizing $M$ will produce a writing of $f$ as a sum of squares. To have three squares, we need that the rank of $M$ is at most $3$. So now the problem is to find a positive semidefinite matrix $M$ of rank at most $3$ and satisfying the equality.

orangeskid
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