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If $ x + y + 2z = 4$ and $ x^2 + y^2 + 2z^2 = 20,$ for the real quantities x, y and z.

How many integral values can z take?

Squaring 1st Equation doesn't help .

Does Arithmetic Mean >= Geometric Mean can be applied?

Thanks in advance.

3 Answers 3

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EDIT: Originally I was trying to solve the problem where all 3 variables are integers (see bellow). The problem posted by OP was asking about integer $z$ and real $x$, $y$, as Théophile pointed out. So here's my second attempt.


Let us have a look first on this problem: For which values of $a$ and $r$ the equalities $\begin{align} x+y&=a\\ x^2+y^2&=r^2 \end{align}$ have a solution?

The answer is: For $a^2\le 2r^2$ or, equivalently, $|a|\le \sqrt{2}r$.

This can be seen geometrically. Draw a picture, the point on circle with radius $r$, which maximizes $x+y$, the point in the direction of the vector $(1,1)$, i.e., the point $(\sqrt{2}r/2,\sqrt{2}r/2)$.

Or simply by solving quadratic equation:
$x^2+(a-x)^2=r^2$
$2x^2-2ax+a^2-r^2=0$
The discriminant of this quadratic equation is
$D=4(2r^2-a^2)$
i.e. it has a real solution if and only if $2r^2-a^2\ge 0$, i.e. $a^2\le 2r^2$.


In our case we have $\begin{align} x+y&=4-2z\\ x^2+y^2&=20-2z^2 \end{align}$ i.e. $a=4-2z$ and $r^2=20-2z^2$. So $a^2\le 2r^2$ is equivalent to
$(4-2z)^2 \le 2(20-2z^2)$
$(2-z)^2 \le 10-z^2$
$z^2-4z+4 \le 10-z^2$
$2z^2-4z-6 \le 0$
$z^2-2z-3 \le 0$
$(z-3)(z+1) \le 0$
The solution of the above inequality is $z\in\langle-1,3\rangle$.


EDIT: The rest of this post is my original solution, where I assumed that all 3 variables should be integers.

From the first equation you get $x=4-y-2z$ and $x^2=16+y^2+4z^2-8y-16z+4yz$. By plugging this into the second equation you get $\begin{align} 16+y^2+4z^2-8y-16z+4yz+y^2+2z^2&=20\\ 2y^2+6z^2-8y-16z+4yz&=4\\ y^2+3z^2-4y-8z+2yz&=2\\ (y+z-2)^2+2z^2-4z&=6\\ (y+z-2)^2+2(z-1)^2&=8 \end{align}$

So now we have a new equation $a^2+2b^2=8$. By trial and error we can find all integer solutions. (We only need to try $a=0,\pm 1,\pm 2$.) We get $a=0$, $b=\pm 2$.

Thus $z=1\pm2$, i.e. $z=-1$ or $z=3$. For each $z$ we get $y=2-z$ and from the first equation we can calculate $x$.

So we have:
$z=-1$, $y=3$, $x=3$
$z=3$, $y=-1$, $x=-1$


Another approach would be going through representations of $20$ as sum of four squares and trying all permutations and possibilities for signs. (I guess this still can be done by hand.) We only want representations $20=x^2+y^2+z^2+w^2$, where $x+y+z+w=4$ and $z=w$.

We have $20=(\pm1)^2+(\pm1)^2+(\pm3)^2+(\pm3)^2$
$20=0^2+0^2+(\pm2)^2+(\pm4)^2$
See wolframalpha.


At the moment I don't see where AM-GM inequality can be used here. From AM-GM we get $\frac{x^2+y^2+z^2+z^2}4 \ge \left(\frac{x+y+z+z}4\right)^2,$ which in this case is $\frac{20}4\ge 1$, which is true. So perhaps if the problem would ask about numbers different from $20$ and $4$, we could show impossibility of (real) solution in this is way, but this is not the case here.

Note also that the AM-GM inequality in this form gives us information only above solutions such that $x,y,z\ge0$.

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    I guess the comment about $(2-\sqrt6,2+sqrt6,0}$ was intended to my original post, where I supposed that $x$, $y$ should also be integers. I've posted a corrected solutions, where I get $-1\le z\le 3$, which includes this possibility.2012-03-26
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The second equation restricts the values of all of the variables, since $x^2$, $y^2$, and $z^2$ must all be non-negative. The closer $x$ and $y$ are to $0$, the greater $z$ can be in absolute value. You can first find a range of real numbers that $z$ is restricted to, then find all integers within that range, and then see whether there are $x$ and $y$ that will satisfy the resulting system.

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$x+y=4-2z \,.$ $x^2+y^2=20-2z^2 \,.$

Then

$(x-y)^2 =2(x^2+y^2) -(x+y)^2 =2*(20-2z^2)-(4-2z)^2 \,$

Now two observations solve the problem:

1)$(x-y)^2 \geq 0 \Rightarrow 2*(20-2z^2)-(4-2z)^2 \geq 0 \,.$

2) If $2*(20-2z^2)-(4-2z)^2 \geq 0$ then the system $x+y=4-2z \,.$ $x-y=\sqrt{2*(20-2z^2)-(4-2z)^2 } \,.$ has solution.

This shows that the system has solution if and only if

$2*(20-2z^2)-(4-2z)^2 \geq 0$

Now all you need to do is figure all the integer solutions of the quadratic inequality:

$24+16z-8z^2 \geq 0 $