Help with a specific Diophantine Equation

Question

I was wondering if anybody knew a way to show that the Diophantine equation given by:

$a^3 + b^3 + c^3 - 3abc =1$

Only has solutions $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$.

The only idea I have had is to consider it modulo $3$, in which case $a^3 =a$ (Fermat's Little Theorem). So then

$a+b+c=1 ($mod $3)$.

But this didn't prove to be very helpful.

Edit: For those interested, this shows that the permutation matrix given by

$P =\left( \begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} \right)$

Does not commute with any $3 \times 3$ matrices which have determinant $1$ and integer entries. The equation $$PA = AP$$ implies that $A= \left( \begin{matrix} a & b & c \\ c & a & b \\ b & c & a \end{matrix} \right),$ and so the determinant equation is $a^3 + b^3 + c^3 -3abc =1$.

are the $a,b,c$ positive naturals? – Dr. Sonnhard Graubner Nov 09 '17 at 19:27 — Dr. Sonnhard Graubner, Nov 09 '17 at 19:27
Michael has gave a better solution than my idea was – Dr. Sonnhard Graubner Nov 09 '17 at 19:29 — Dr. Sonnhard Graubner, Nov 09 '17 at 19:29

Michael Rozenberg · Accepted Answer · 2017-11-09T19:48:26.547

3

Since $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$ and $$a^2+b^2+c^2-ab-ac-bc=\frac{1}{2}\sum_{cyc}(a-b)^2\geq0,$$ we obtain $$a+b+c=1$$ and after substitution $c=1-a-b$ we obtain $$a^2+ab+b^2=a+b$$ or $$a^2+(b-1)a+b^2-b=0,$$ which gives $$(b-1)^2-4(b^2-b)\geq0$$ or $$-\frac{1}{3}\leq b\leq1,$$ which gives $$b\in\{0,1\}.$$ By the same way we can get that $a\in\{0,1\}$ and with $a+b+c=1$ it gives the answer.

edited Nov 09 '17 at 19:48

answered Nov 09 '17 at 19:27

Michael Rozenberg

194,933

So this implies that (a+b+c) and the other factor both have to be one or negative one? And we know that (a+b+c) has to be one (it can't be negative)? – Nov 09 '17 at 19:30
@Jbag1212 Are our variables integers or they are non-negatives? – Michael Rozenberg Nov 09 '17 at 19:33
For the problem I am solving at the moment I only need non-negatives, however in the future it might be useful to know if this works for the negatives as well. – Nov 09 '17 at 19:34
@Jbag1212 If they are non-negatives we obtain $a+b+c=1$, which gives the answer. – Michael Rozenberg Nov 09 '17 at 19:36
@Jbag1212 Show that $a^2+b^2+c^2-ab-ac-bc\ge0$ so that we need not worry about the possibility that $a+b+c=-1$. – Angina Seng Nov 09 '17 at 19:41
@Jbag1212 I have a solution for all integer variables. If you want I am ready to post it. – Michael Rozenberg Nov 09 '17 at 19:43
@Micheal That would be great. – Nov 09 '17 at 19:43
@Jbag1212 I added something. See now. – Michael Rozenberg Nov 09 '17 at 19:54

Will Jagy · Answer 2 · 2017-11-09T19:51:20.840

1

The next ingredient is that $$ a^2 + b^2 + c^2 - ab - ac - bc = \frac{1}{2} \left( (a-b)^2 + (b-c)^2 + (c-a)^2 \right). $$ This is non-negative and can only be $1$ for the two patterns $(t,t,t+1)$ or $(t, t+1, t+1)$ in order. If, however, $t <0,$ the sum $a+b+c$ becomes negative. So, what remains (see comment by Shark) is $t \geq 0$ and either $3t+1=1$ or $3t+2 = 1.$ As $t$ is an integer, it must be the first pattern, with $t=0,$ giving $(0,0,1)$ when put in order.

edited Nov 09 '17 at 19:51

answered Nov 09 '17 at 19:45

Will Jagy

139,541

1

Even more obviously, as one must have $a+b+c=1$ then one also has $3t+1=1$. – Angina Seng Nov 09 '17 at 19:47

score 0 · Answer 3 · answered Nov 10 '17 at 04:58

Strangely enough, the solution is finite.

for the equation:

$$X^3+Y^3+Z^3-3XYZ=q=ab$$

If it is possible to decompose the coefficient as follows: $4b=k^2+3t^2$

Then the solutions are of the form:

$$X=\frac{1}{6}(2a-3t\pm{k})$$

$$Y=\frac{1}{6}(2a+3t\pm{k})$$

$$Z=\frac{1}{3}(a\mp{k})$$

Thought the solution is determined by the equation Pell, but when calculating the sign was a mistake. There's no difference, but the amount should be. Therefore, the number of solutions of course.

Help with a specific Diophantine Equation

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