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"solve this diophantine equation: $a^3-b^3 = ab+61$"

For Diophantine Equations I have always had the question in the form $ax+by = c$ and solved it via Extended Euclidean Algorithm. However, this question is not formatted in such a way.

What I am thinking is that we move $ab$ to the LHS so that we can have a single constant on the RHS. Maybe then we could expand? I'm not sure. Hints/Advice is what I am looking for.

Maybe we can substitute $a$ and $b$ with something else of a form I am more comfortable with? I'm not really sure.

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    $\begingroup$ Solving 3rd degree diophantine equations is, in general, MUCH harder than 1st degree equations like $ax+by=c$. Occasionally, there's a trick, but for the most part this is not "elementary-number-theory". $\endgroup$ – Gerry Myerson Oct 6 '13 at 6:49
  • $\begingroup$ @GerryMyerson Ah Sorry, my mistake $\endgroup$ – Ozera Oct 6 '13 at 16:08
  • $\begingroup$ Unless the answers below are wrong, this did indeed turn out to be fairly elementary. $\endgroup$ – Kieren MacMillan Oct 23 '13 at 19:13
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Finding all integer solutions to the equation $a^3-b^3=ab+61$ is equal to solving the diophantine equation $a^3-ab-b^3=61$, and placing $-b$ instead of $b$ gives $a^3+ab+b^3=61$.

Since the equation above is symmetric ( that is, the value doesn't changes even if we swap the position of the variables a and b), it can be expressed with the sum and product of a and b. Let c=a+b and d=ab. Then the quadratic equation $t^2-ct+d$ has two zeros, namely a and b. Thus, the discriminant $D=c^2-4d≥0$, which yields $d\le \frac {c^2}{4}$ ( and of course $c, d \in\mathbb Z$). In addition, $a^3+ab+b^3=61 \iff a^3+b^3+ab=61$

$\iff (a+b)^3-3(a+b)ab+ab=61$

$\iff c^3-3cd+d=61$

$\iff d=(c^3-61)/(3c-1)$

Since $d \in\mathbb Z$, $27d = \frac{27c^3-1647}{3c-1} \in\mathbb Z$.

But then, $\frac{1646}{3c-1} \in\mathbb Z$, because $\frac{27c^3-1647}{3c-1} = 9c^2+3c+1\in\mathbb Z$. This means that $3c-1$ must divide $1646=2\cdot 823$ (823 is a prime). So $3c-1$ can be 1, 2, 823, 1646, -1, -2, -823, or -1646. Then $3c$ can be 2, 3, 824, 1647, 0, -1, -822, or -1645. Since c is an integer, the only possible values for c are c=0, 1, 549. This gives us the pair $(c,d) = (0,61), (1, -30), (549, 100528)$, and the only pair that $d\le \frac{c^2}{4}$ holds is $(1, -30)$. Thus, $c=1$ and $d=-30$, and it follows that a and b are the two zeros of the quadratic equation $x^2-x+30=0$. We now can conclude that the only solutions to the Diophantine equation $a^3+ab+b^3=61$ are $(a, b) = (6, -5), (-5, 6)$.

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$$(a-b)(a^2+ab+b^2)=ab+61$$ $$\implies a^2+ab+b^2 \le \mid ab+61\mid$$ The negative case inside the absolute value readily leads to contradiction, so now we have $$a^2+b^2\le61$$ These are finitely many cases to check(having even less if we note that $a\ge b$ and that $(a,b)=1$ (why?)), so we end up with $a=6,b=5;a=-5,b=-6$.

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    $\begingroup$ Lovely proof! It's amazing what a little clever insight (like $a^2+ab+b^2 \le \lvert ab+61 \rvert$) will do to make a difficult problem much easier. $\endgroup$ – Kieren MacMillan Oct 6 '13 at 12:51
  • $\begingroup$ @KierenMacMillan Thank you!, it was a moment of inusual(to my standards) inspiration. $\endgroup$ – chubakueno Oct 6 '13 at 15:25
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    $\begingroup$ oh wow very nice. Simply observing $a^2 + ab + b^2 \leq |ab +61|$ does so much! $\endgroup$ – Ozera Oct 6 '13 at 16:11
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    $\begingroup$ @Ozera My poor experience in elliptic curves and diophantine cubics has taught me that size and modular considerations are powerful :) $\endgroup$ – chubakueno Oct 6 '13 at 17:26

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