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Are there known any other simple nontrivial polynomials in two or more variables that can be factored like the one in Sophie Germain's identity? $$x^4+4y^4=(x^2-2 x y+2 y^2) (x^2+2 x y+2 y^2)$$

Besides the well known (for $n \in \mathbb {Z} ^{+}$): $$x^n-y^n=(x-y) \sum _{i=1}^n x^{i-1} y^{n-i}$$ $$x^{2 n-1}+y^{2 n-1}=(x+y) \sum _{i=1}^{2 n-1} (-1)^{i-1} x^{i-1} y^{2 n-1-i}$$ And of course besides trivial ones like $5 x^2+x y^2=x (5 x+y^2)$, where all terms share a common factor.

By simple polynomial I mean a polynomial that has few terms. The less terms the better - ideally only two terms like in the Germain's polynomial or in the known ones.

azerbajdzan
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3 Answers3

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More formulas similar to Sophie Germain's identity can be derived from Aurifeuillean factorizations such as

$$x^6+27y^6=(x^2+3y^2)(x^2-3xy+3y^2)(x^2+3xy+3y^2),$$ $$x^{10}-3125y^{10}=(x^2-5y^2)(x^4-5x^3 y+15x^2 y^2-25x y^3+25y^4)(x^4+5x^3 y+15x^2 y^2+25x y^3+25y^4).$$

Junka
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  • Indeed. I want to add a link to my favorite source on Aurifeuillian stuff. Paul Garrett actually visits the site regularly. – Jyrki Lahtonen Oct 12 '19 at 16:43
  • Do you have similar examples also for polynomials in three variables? Or in two variables but each variable has different exponent? – azerbajdzan Oct 12 '19 at 16:44
  • @azerbajdzan Sophie Germain's factorization appears occasionally in contests. Here's a contest math application of an Aurifeuillian factorization related to the fifth cyclotomic. Sorry about blowing my own trumpet a bit. – Jyrki Lahtonen Oct 12 '19 at 16:48
  • @Junka: I accepted your answer too early. Your examples are trivial. They reduce into $x^6+27 y^6=(x^2)^3+(3 y^2)^3$ and $x^{10} - 3125 y^{10} = (x^2)^5-(5 y^2)^5$ which are the "well know" examples I already provided in my question. – azerbajdzan Oct 12 '19 at 17:17
  • @Jyrki Lahtonen: So can you provide (using Aurifeuillean factorization) an explicit example of a polynomial that cannot be transformed to cases I already mentioned in my original question? – azerbajdzan Oct 12 '19 at 17:24
  • The one I used in the linked question is Junka's second factorization with the obvious values plugged into $x$ and $y$. I need to think harder for something else. Don't hold your breath! – Jyrki Lahtonen Oct 12 '19 at 17:47
  • @JyrkiLahtonen I put one of my quirky examples. – Will Jagy Oct 12 '19 at 19:24
  • @azerbajdzan And I would not call the extra factorization (missing from your examples) "trivial". The left hand sides are in your collection, but the right hand sides are the point here, as there is more factorization taking place. Furthermore, they are not available for all choices $n$, and crucially need the extra constant factors. – Jyrki Lahtonen Oct 15 '19 at 17:07
  • @Jyrki Lahtonen: Sorry but maybe I do not understand your point here - maybe you could explain it to me in more detail. In my eyes they are trivial. Once you know the formulas I provided in my question (which are themselves not trivial of course and of course I am not inventor of them) - but once you know the formulas you can make such examples like Junka's ones as much as you want: $$x^{22}+2048 y^{22}=(x^2+2 y^2) p_1(x,y)$$ $$8 x^{12}+125 y^{30}=\left(2 x^4+5 y^{10}\right) \left(4 x^8-10 x^4 y^{10}+25 y^{20}\right)$$ $$x^{91}+19487171 y^{119}=(x^{13}+11 y^{17}) p_2(x,y)$$ – azerbajdzan Oct 15 '19 at 20:24
  • @Jyrki Lahtonen: ...and without knowing anything about Aurifeuillian factorization. – azerbajdzan Oct 15 '19 at 20:26
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    @azerbajdzan All the factorizations in your list only have two factors on the right hand side, but Junka's has three. In my opinion that is the key difference. – Jyrki Lahtonen Oct 15 '19 at 20:33
  • @Jyrki Lahtonen: Ahh... now I see...he/she did not stress out that fact. This is interesting property, but what it has to do with Sophie Germain's identity? It has only two factors too (over integers). So factoring something into three factors does not make it more "like Germain's identity" than something that factors only into two factors. I found Dietrich Burde's example $x^3+y^3+z^3-3xyz$ more similar to Germain's identity. Only the term $3 x y z$ makes it, say "less attractive" than Germain's. But it is still simple polynomial and non-trivial regarding factorization. – azerbajdzan Oct 15 '19 at 20:48
  • @Jyrki Lahtonen: Can you provide your own example similar to Junka's ones but such that only nontrivial factorization is possible and none factorization according to "my" formulas? – azerbajdzan Oct 15 '19 at 20:57
  • Ok. I also see that you are looking for factorizations of very simple polynomials in more variables. The Aurifeuillea stuff is actually univariate (or bivariate after a homogenuzation). I'm sure something is know there, but I can't recall anything right now. – Jyrki Lahtonen Oct 15 '19 at 21:02
  • @Jyrki Lahtonen: If I remove the trivial factor $x^{2} + 3 y^{2}$ from junka's example $x^{6} + 27 y^{6}$ then we get $x^4-3 x^2 y^2+9 y^4=\left(x^2-3 x y+3 y^2\right) \left(x^2+3 x y+3 y^2\right)$. And in this shape it is far more like Germain's identity. :-) But doing the same for the second example we get polynomial that has 5 terms and I would not call it simple anymore. – azerbajdzan Oct 15 '19 at 21:05
  • @Jyrki Lahtonen: I was wondering whether there is a family of formulas where the Sophie Germain's identity belongs. Like $x^3+y^3$, $x^4+y^4$ and so on belongs to one factoring family. But is seems that Sophie Germain's identity is one of a kind. It does not belong to any family of formulas. – azerbajdzan Oct 15 '19 at 21:12
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The roots of $\eta^3 - 3 \eta - 1 = 0$ are $$ A = 2 \cos \left( \frac{7 \pi}{9} \right) \approx -1.532 \; \; \; , B = 2 \cos \left( \frac{5 \pi}{9} \right) \approx -0.347 \; \; \; , C = 2 \cos \left( \frac{ \pi}{9} \right) \approx 1.879 \; \; \; . $$ We get identity $$ \color{magenta}{ (x+y+z)^3 - 9 \left( x^2 y + y^2 z + z^2 x \right) = (Ax+By+Cz)(Bx+Cy+Az)(Cx+Ay+Bz) } $$

There is a theorem: given a ternary cubic $f, \; $ write out the Hessian matrix $H.$ The entries of $H$ are homogeneous linear in the three variables. As a result, the determinant $\Delta$ of $H$ is once again a ternary cubic. If $\Delta$ is a constant multiple of $f, \; $ then $f$ factors as the product of three homogeneous linear factors, although it might be necessary to allow complex coefficients. This all applies to Dietrich's example, where he does need complex coefficients.

Will Jagy
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For example, $$ x^3+y^3+z^3-3xyz=(x+y+z)(x+\omega y+\omega^2z)(x+\omega^2y+\omega z), $$ for a third root of unity $\omega$.

Dietrich Burde
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  • A good one. Which can be written as $(x+y+z) \left(x^2-x y-x z+y^2-y z+z^2\right)$ if we are working only over integers. Any other? – azerbajdzan Oct 12 '19 at 15:44
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    Yes, for example $x^4 + y^4 + (x + y)^4 = (x^2 + xy + y^2)^2$. – Dietrich Burde Oct 12 '19 at 15:48
  • A nice one too! You probably missed "2" on the right side $x^4+y^4+(x+y)^4=2 (x^2+x y+y^2)^2$. But if we expand left side we get $2 x^4 + 4 x^3 y + 6 x^2 y^2 + 4 x y^3 + 2 y^4$ which has more terms. But I like it anyway. What about a crazy one like $3 x^7-17 y^4+105 z^6$ (which is of course irreducible) but is there any chance for such a crazy example? – azerbajdzan Oct 12 '19 at 15:55