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My question is: Solve simultaneously:(anwers are in integers)

$$\begin{align} y^3 - 9x^2 + 27x - 27 &= 0 \\ z^3-9y^2+27y-27 &= 0 \\ x^3-9z^2+27z-27 &= 0 \end{align}$$

Any hints to solve this question would be greatly appreciated.

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2  
At least $(3,3,3)$ should work. –  TMM Jun 2 '12 at 14:20
    
@TMM:But how can we solve this simultaneously? –  mgh Jun 2 '12 at 14:21
    
If you use the newton-method for systems of equations, that gives you a chance to get an approximation at least. –  Peter Sheldrick Jun 2 '12 at 14:29
    
@Peter Sheldrick I have no idea about that.Plz can you tell me how can i solve this question simultaneously like by substitution and all –  mgh Jun 2 '12 at 14:45
4  
-1 for responding to each comment by simply repeating your question without any indication that you have actually read the comment. –  Henning Makholm Jun 2 '12 at 15:17

1 Answer 1

up vote 9 down vote accepted

Edit: It turns out that it is only integer solutions that are being asked about. Then there is a simpler approach that involves play with inequalities. Below is the original answer, followed by a solution of the integer version of the problem.

A start: Note that the first equation can be rewritten as $$y^3+(x-3)^3=x^3.$$

Similarly, the other two can be rewritten as $$z^3+(y-3)^3=y^3$$ and $$x^3+(z-3)^3=z^3.$$ Add up, cancel. We get $$(x-3)^3+(y-3)^3+(z-3)^3=0.$$ If you are interested in integer or rational solutions, the problem was long ago settled by Euler, and (perhaps) earlier by Fermat. If you want to examine the system over the reals, there is more work to do.

Integer solutions, another approach: First we note that by the case $n=3$ of Fermat's Last Theorem, any solution of $a^3+b^3=c^3$ in integers (with some possibly negative) has at least one of $a$, $b$, or $c$ equal to $0$. The case $n=3$ of FLT has a relatively elementary, though by no means simple proof that goes back to Euler. A proof may even have been given by Fermat, though there is documentary evidence only for the case $n=4$. Using FLT, we can easily classify all the solutions of our system.

Now we show how to solve the problem without using the case $n=3$ of FLT. First we do a simplification that makes the coefficients smaller.

From the first equation, it is clear that $y^3$ is divisible by $3$, so $y$ must be. Let $y=3v$. Similarly, $z$ is divisible by $3$, say $z=3w$, and $x$ is divisible by $3$, say $x=3u$. After the substitutions have been done, we arrive at the equations $$\begin{align} v^3 &=3u^2 -3u+1 \\ w^3 &=3v^2 -3v+1\\ u^3 &=3w^2 -3w+1 \end{align}$$ Note that for any real number $t$, we have $3t^2-3t+1\ge \frac{1}{4}$. So since we are looking for integer solutions, all of $u$, $v$, and $w$ are positive integers.

If $t$ is a positive integer, then $3t^2-3t+1 \lt 3t^2$. So from our system of equations we conclude that $$v^3 \lt 3u^2, \qquad w^3 \lt 3v^2,\qquad u^3 \lt 3w^2.$$ From the first inequality we get $v^{27}\lt 3^9 u^{18}$. But by the third inequality, $u^{18}\lt 3^6 w^{12}$. But by the second inequality $w^{12} \lt 3^4 v^8$. Putting things together, we get $$v^{27}\lt 3^{19}v^8,$$ from which it follows that $v^{19} \lt 3^{19}$, or more simply $v\lt 3$. By symmetry we also have $u \lt 3$ and $w \lt 3$. Thus each of our variables $u$, $v$, and $w$ must be $1$ or $2$, and now we are essentially finished.

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I got till the step where we add up and cancel but i did not get the one you wrote in bracket –  mgh Jun 2 '12 at 15:40
    
You may be referring to something I edited out, in order not to give away too much. If we let $u=x-3$, $v=y-3$, $w=z-3$, then our equation becomes the simpler-looking $u^3+v^3+w^3=0$. –  André Nicolas Jun 2 '12 at 15:44
    
i want the answer in integers –  mgh Jun 2 '12 at 15:52
2  
In your course, the Fermat equation $a^n+b^n=c^n$ must have been mentioned. The equation I reached is a close relative of the Fermat equation for $n=3$. Bu there is another way that I will give a hint for. –  André Nicolas Jun 2 '12 at 15:55
    
ok thanks I will wait for the other hint –  mgh Jun 2 '12 at 16:10

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