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Can someone show me how to find all solutions in positive integers to the diophantine equation: $$x^3 + y^3 = 35$$ I know how to do it algebraically, but I want to know how you solve it in number theory.

My Algebraic Approach: $x^3 + y^3 = (x + y)\cdot (x^2 - xy + y^2) = 35$.

The only integer factors of $35$ are $(1, 35)$ or $(5, 7)$. There are no integers $x$ and $y$ that add to $1$, so $x + y = 5$ or $7$.

Using $x + y = 5$ we get that $y = 5 - x$, so $y^3 = 125 - 75x + 15x^2 - x^3$ so $x^3 + y^3 = 15x^2 - 75x + 125 = 35$ or $$15x^2 - 75x + 90 = 0$$ $$x^2 - 5x + 6 = 0$$ $$(x - 3)(x - 2) = 0$$ So $x = 2$ or $x = 3$ Thus $y = 3$ or $y = 2$ respectively.

Using $x + y = 7$ we get that $y = 7 - x$, so $y^3 = 343 - 147x + 21x^2 - x^3$ so $$x^3 + y^3 = 21x^2 - 147x + 343 = 35$$ $$21x^2 - 147x + 308 = 0$$ $$3x^2 - 21x + 44 = 0$$ or $$x = \frac{21 \pm \sqrt{-87}}{6}$$ Since x is complex, this can't be a solution.

So $(x, y) = (2, 3)$ or $(x, y) = (3, 2)$

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  • $\begingroup$ what is your algebraic way? Please share so that help can be provided accordingly. $\endgroup$ – Anurag A Feb 28 '15 at 1:33
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    $\begingroup$ $4^3=64$, so $x$ and $y$ are both between $1$ and $3$, inclusive. Thus, an exhaustive search is far quicker than trying to be smart and slick. $\endgroup$ – Henning Makholm Feb 28 '15 at 1:34
  • $\begingroup$ It's $x^2\color{red}-xy+y^2$. $\endgroup$ – Lucian Feb 28 '15 at 2:21
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    $\begingroup$ @rachel: Your argument was number-theoretic, since you used the factorization of $35$. It was well done, but for a specific small number like $35$, it is an unnecessarily complicated approach. $\endgroup$ – André Nicolas Feb 28 '15 at 2:39
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    $\begingroup$ Now find all solutions in positive integers to $x^3+y^3=1729$, and then tell G H Hardy about it. $\endgroup$ – Gerry Myerson Feb 28 '15 at 3:12
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You know that $x$ and $y$ cannot both be greater than or equal to $3$, because this would give you too large of a sum. And $x$ and $y$ must both be less than $4$ by similar reasoning. So in this case, that's probably the best way to approach the problem: Just try $1,2,3$ for $x$ and $y$, where they are not both $3$. And note you can assume $x \leq y$ by symmetry, and modify your answer after the fact based on symmetry and the solutions you get.

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To bound all solutions in integers (not just to positive integers) to $x^3+y^3 = n$:

First, using the factorization $x^3+y^3 =(x+y)(x^2-xy+y^2) $, we get possible values for $x+y$ since $(x+y) | n$. Then, since $x^2-xy+y^2 =(x+y)^2-3xy $, we get possible values for $xy$ ($xy =\dfrac{(x+y)^2-n/(x+y)}{3} $) and this gives $x$ and $y$.

Note (added after a comment): If $a=x+y$ and $b=xy$ are known, then, since $(r-x)(r-y) =r^2-r(x+y)+xy =r^2-ar+b $, $x$ and $y$ are the roots of $r^2-ar+b$. So use the quadratic formula: $r =\dfrac{a\pm \sqrt{a^2-4b}}{2} $.

(Additional stuff to make the solution explicit)

We have $a = x+y$ (the divisor of $n$), $b = xy = ((x+y)^2−n/(x+y))/3=(a^2−n/a)/3$. Then the discriminant $d =a^2-4b =a^2-4(a^2−n/a)/3 =(4n/a-a^2)/3 =(4n-a^3)/(3a) $ and the roots are $\dfrac{a\pm \sqrt{(4n-a^3)/(3a)}}{2}$. If we adopt the convention that $x < y$, then $x = \dfrac{a- \sqrt{(4n-a^3)/(3a)}}{2}$ and $y = \dfrac{a + \sqrt{(4n-a^3)/(3a)}}{2}$ . For this to work, $(4n-a^3)/(3a)$ must be a perfect square.

Some elementary bounds on $x$ and $y$:

If $0 < x \le y$, $\sqrt[3]{n/2} < y \lt \sqrt[3]{n} $.

If $x < 0 < y$, since $x^2-xy+y^2 =x^2-xy+(y/2)^2+3y^2/4 =(x-y/2)^2+3y^2/4 > 3y^2/4 $, $x+y > 0$ or $y > -x$.

This can be improved to $x^2-xy+y^2 =x(x-y)+y^2 =(-x)(y-x)+y^2 > y^2 $ and $x^2-xy+y^2 <3y^2 $.

For every divisor $d$ of $n$, $d y^2 < n < 3dy^2$, so $\sqrt{n/(3d)} < y < \sqrt{n/d}$.

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  • $\begingroup$ And then how do I find x and y exactly? $\endgroup$ – rachel Feb 28 '15 at 3:44
  • $\begingroup$ I added an explanation. Essentially, use the quadratic formula. $\endgroup$ – marty cohen Feb 28 '15 at 3:54
  • $\begingroup$ In this case what is a and b $\endgroup$ – rachel Feb 28 '15 at 4:00
  • $\begingroup$ I don't understand what I plug into the quadratic formula for a and b. $\endgroup$ – rachel Feb 28 '15 at 4:02
  • $\begingroup$ a = x+y (the divisor of n), b = xy = $((x+y)^2-n/(x+y))/3 = (a^2-n/a)/2$. $\endgroup$ – marty cohen Feb 28 '15 at 5:03
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COMMENT.-You have solved easily the problem algebraically and you want to know how solve it in number theory. Well your curve is an elliptic curve of rank $1$ whose generator is the point $(3,2)$ which is the only point of integer coordinates and it is determined using your algebraic way. For the rest it has an infinity of rational points whose set is dense in the curve and that can be calculated by the method of chords and tangents. That is what number theory in connection with elliptic curves tells us about this problem (algebraic number theory deals with the other non-rational algebraic points, i.e. non trascendental, of the curve).

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