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If $x,y\in\mathbb{Z^+}$, then find all the integral solutions to:

$$x^6-y^6+3x^4y-3y^4x+y^3+3x^2+3x+1=0$$

I tried solving this question for an hour but still couldn't get it. I tried mod reduction and factorization, but couldn't do it. I welcome any approach (elementary or non elementary) for this question.

Any help will be appreciated.
Thanks in advance.

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    $\begingroup$ Please make titles informative. $\endgroup$
    – Pedro
    Jan 16, 2015 at 15:58

3 Answers 3

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Given: $x^6-y^6+3x^4y-3y^4x+y^3+3x^2+3x+1=0$

$\implies(x+1)^3+(x^2+y)^3=(x+y^2)^3$

By Fermat's Last Theorem, since $x,y\in\mathbb{Z}^+$ , this equation has no solution.

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    $\begingroup$ +1 interesting. I never expect this simple equation can be solved using FLT. $\endgroup$ Nov 30, 2014 at 14:15
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    $\begingroup$ Kids these days... back in my day, FTL wasn't proven, ergo this solution would not have been accepted. $\endgroup$
    – Michael
    Nov 30, 2014 at 17:44
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    $\begingroup$ @Michael The case for cubes was, though. (Unless you're a few hundred years old) $\endgroup$ Nov 30, 2014 at 18:31
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    $\begingroup$ I have discovered a truly marvelous proof of this, which this comment is too short to contain. $\endgroup$ Dec 1, 2014 at 4:46
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    $\begingroup$ @TavianBarnes Actually me and MathGod both go to the same school and both of us have received the worksheet which contains this question. The correct question is the one after the edit and I had missed the $y^3$ term in the earlier one. MathGod might have already solved this question from the worksheet, before this question was put on this website, and thus answered according to the question in the worksheet (with the $y^3$ term present). I apologize for any miscommunication caused. $\endgroup$
    – Henry
    Dec 2, 2014 at 8:06
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Hint:

You can rearrange this as $$(x+1)^3-x^3=y^6-x^6+3xy(y^3-x^3)=(y^3-x^3)(y^3+3xy+x^3)$$

And note that $x,y\in \mathbb Z^+$

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  • $\begingroup$ Sorry, I didn't get it. Can you please elaborate? $\endgroup$
    – Henry
    Nov 30, 2014 at 14:15
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    $\begingroup$ @Samurai Since LHS is positive, RHS need to be positive. As a result, $y \ge x + 1$ and RHS is at least 5 times of LHS, a contradiction! $\endgroup$ Nov 30, 2014 at 14:22
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I think it can be done directly, without reducing to FLT. A bit of algrebra shows that $|x^6 - y^6|$ dominates the lower order terms for all except a very small finite collection of $x$ and $y$ (which can be checked by hand):

First, it's easy to see that any solution must have $y \gt x \gt 0$, so write $y = x + a$, with $a \gt 0$ also.

Now substitute and expand everything. I get, $$ y^6 - x^6 = 6ax^5 + 15a^2x^4 + 20a^3x^3 + 15 a^4x^2 + 6a^5x + a^6 $$ while, if my pen-and-paper calculation is correct, $$ 3x^4y - 3xy^4 + 3x^2 + 3x + 1 = -(3a^4-1)x - (12 a^3 - 1) x^2 - 18 a^2 x^3 - 9 a x^4 + 1 $$ Comparing terms in the two expressions above, we see that the sixth order terms are strictly larger (in absolute value) than lower order term involving the same power of $a$, (except for very small $a,x$). E.g., $$ 18 a^2 x^3 < 15 a^2 x^4 \quad \textrm{unless} \quad a,x < 2 $$ $$ |(3a^4 - 1)x| < 15 a^4 x^2 \quad \textrm{if} \quad a,x \ge 1 $$ and also, $a^6 > 1$ for $a > 1$.

All the sixth order terms have the same sign. Then we have that the sum of the sixth order terms is strictly larger than the absolute sum of the lower order terms. The remaining cases (I think just $a,x \in {1,2}$) can be checked by hand.

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  • $\begingroup$ Even assuming I got the algebra correct above, I still prefer the reduction to FLT :) ... $\endgroup$ Dec 1, 2014 at 5:04
  • $\begingroup$ Didn't Mark Bennet give an elegant solution that doesn't involve anything like FLT at all? $\endgroup$
    – user21820
    Dec 1, 2014 at 5:07
  • $\begingroup$ Ah yes I was busy typing in my non-elegant solution and didn't notice his. I should read before I post. $\endgroup$ Dec 1, 2014 at 5:10
  • $\begingroup$ Ah I see. No problem, and anyway welcome to Math SE! I hope you find other interesting problems here, or come with your own! $\endgroup$
    – user21820
    Dec 1, 2014 at 5:13

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