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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.

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3 Answers

up vote 3 down vote accepted

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:
The discriminant of this quadratic equation is
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 $20=(\pm1)^2+(\pm1)^2+(\pm3)^2+(\pm3)^2$
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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According to the problem, though, only $z$ need be an integer; $x$ and $y$ are real. The solution is therefore much more simple: choose any $z \in \mathbb{Z}$ such that $2z^2 \leq 20$ (i.e., between $-3$ and $3$), and it is straightforward to find $x$ and $y$ that satisfy the system. In other words, the answer is $7$, since $z \in \{-3, -2, -1, 0, 1, 2, 3\}$. –  Théophile Mar 25 '12 at 16:14
@Theophile, you're right. I should have read the problem more carefully. –  Martin Sleziak Mar 25 '12 at 16:34
@Martin thanks so much. –  vikiiii Mar 25 '12 at 17:13
@Martin ya .you are right.Theophile answer includes -3 and -2 also.if we conside z=-3 , then according to the 1st condition x+y should be equal to 10, because of this any combination of x and y will violate the 2nd equation.therefore -3 cant be answer .and similarly -2. –  vikiiii Mar 25 '12 at 17:24
A user who does not have enough points to comment wishes to add: "It is incorrect. Since for (x,y,z) = (2+sqroot(6), 2-sqroot(6), 0) is also a solution. I cannot comment that's why I write here." –  Zev Chonoles Mar 25 '12 at 21:27
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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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$$x+y=4-2z \,.$$ $$x^2+y^2=20-2z^2 \,.$$


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

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:

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

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