Find all integers $x$, $y$, and $z$ such that $$xyz = 288$$ and $$xy + xz + yz = 144\,.$$
I did this using brute force, where $$288 = 12 \times 24 = 12 \times 6 \times 4$$ and found that these set of integers satisfy the equation. How do I solve this without using brute force?
On
$$\begin{cases}xyz = 288\\xy + xz + yz = 144\end{cases}\overset{Resultant_z}{\implies}288 x + 288 y - 144 x y + x^2 y^2=0\implies$$
$$\Bigl(36 (x y^2 - 72 (y - 2))\Bigr)^2 = 26873856 + 373248 (-72 y) + 1296 (-72 y)^2 + (-72 y)^3$$
magma-code:
V:= [];
S:= IntegralPoints(EllipticCurve([0, 1296, 0, 373248, 26873856]));
for s in S do
y:= s[1]/(-72);
if (y ne 0) and (y eq Floor(y)) then
x:= (s[2]/36+72*(y-2))/y^2;
if x eq Floor(x) then
z:= 288/x/y;
if z eq Floor(z) then
if (x le y) and (y le z) then
V cat:= [[x,y,z]];
end if;
end if;
end if;
end if;
end for;
if #V ge 1 then
for v in V do printf "%o,", v; end for;
printf "\n";
end if;
quit;
with output [ 4, 6, 12 ].
On
We have $288 = 2^53^2.$ Let $x=2^a3^r,$ $y=2^b3^s,$ and $z=2^c3^t$. Then $a+b+c = 5$ and $r+s+t=2$. Since $r,s,$ and $t$ are non-negative integers, one of them must be $0$, say $t=0$. From the equation
$$xy+xz+yz = 144,$$
we see that if a prime number divides any of the variables, it must divide at least one of the others. This forces $r =s=1$ and we must have
$$x=2^a3, y=2^b3, z=2^c.$$
Similarly, if one of the variables is divisible by $8$, then the product of the other two variables is also divisible by $8$, but then $a+b+c\geq 6$, which is too big. This forces $1\leq a,b,c \leq 2.$
So either $c=1$ or $c=2.$ If $c=1$, then $a=b=2$ and $x=y=12$ and $z=2$, which doesn't satisfy the second equation.
If $c=2$, then $a=2, b=1$ (or vv.) and we have $x=12$, $y=6$, $z=4$ which is the only solution.
On
This solution is not using any brute force, not even a simple value trial-and-error, but not using egyptian fractions either.
$xyz = 288 \tag{1}$
$xy + xz + yz = 144 \tag{2}$
As addition & multiplication are applied to all variables, they can be exchanged with each other, so if we prove something for any of them, that is true for any other of them. We have 2 cases, when all variables are positive integers, and when 2 of them are negative.
1. When 2 of them are negative
then let those be $x$ and $y$. From $(2)$ it follows, as $xy > 0$ and $xz < 0$ and $yz < 0$, that $xy > 144$. According to $(1)$ $xy \mid 288$, and this is only possible in the case of $xy > \frac{288}{2}$, if $xy = 288 \Rightarrow z = 1 \tag{3}$
and from this $(2)$ is
$288 + x + y = 144$
$x + y = -144$
as $x$ and $y$ are both negative, so far it is OK.
$x = -144 -y \tag{4}$
Substituting $(3)$ and $(4)$ into $(1)$ we get
$(-144 - y)y = 288$
$y^2 + 144y - 288$
$y_{1,2} = \frac{-144 \pm \sqrt{144^2 + 4 \times 288}}{2}$
Checking the determinant, we get
$144^2 + 4 \times 288 = 144^2 + 8 \times 144 = 144(144 + 8) = 12^2 \times 152$
and as the determinant should be a square and $152$ is not one, $y$ can not have integer solution.
2. When all of them are positive
then from $(2)$ it follows, as $xy > 0$ and $xz > 0$ and $yz > 0$, that $yz < 144 \Rightarrow yz = \frac{144}{A}, A \in\mathbb{Z}^+ \Rightarrow x = 2A \tag{5}$
Based on this writing up $(2)$ we get
$2Ay + 2Az + \frac{144}{A} = 144$
and from this we can get a quadratic for $A$:
$(y + z)A^2 -72A + 72 = 0$
$A_{1,2} = \frac{72 \pm \sqrt{72^2 - 4 \times 72(y + z)}}{2(y + z)} \tag{6}$
Checking the determinant, we get
$72^2 - 4 \times 72(y + z) = 72(72 - 4(y + z)) = 72 \times 4 (18 - (y + z)) \tag{7}$
As the determinant should not be negative, therefore $(y + z) \le 18 \Rightarrow (y + z) = \frac{18}{B}, B \in\mathbb{Z}^+ \tag{8}$
First let's see, if $B \ne 1 \tag{9}$
$(7)$ can be written further with $(8)$ as
$72 \times 4 (18 - (y + z)) = 72 \times 4 (18 - \frac{18}{B}) = 72 \times 4 \times 18 (1 - \frac{1}{B}) = 72^2 (1 - \frac{1}{B})$
and as the determinant should be a square, and $1 - \frac{1}{B} < 1$, therefore $1 - \frac{1}{B} = (\frac{C}{D})^2, C,D \in\mathbb{Z}^+, D > C $, from which
$\frac{B - 1}{B} = \frac{C^2}{D^2}$
i.e. a square of a fraction must be a fraction of squares, which need to be consecutive numbers. And this is not possible, because if $B = x^2, x \in\mathbb{Z}^+ \Rightarrow B - 1 = x^2 - 1 \tag{10}$
and we want $x^2 - 1 = y^2, y \in\mathbb{Z}^+$, then
$x^2 - y^2 = 1$
$(x + y)(x - y) = 1$
and this is only possible for integers, if both terms are 1:
$x + y = 1$
$x - y = 1$
therefore
$x + y = x - y$
$y = -y$
$y = 0 \Rightarrow x = 1$ per above
but $y$ must be $ > 0$ per definition, otherwise per $(10)$ there is a contradiction for $B$ per $(9)$, as $x = 1 = B$.
Therefore let's see, if $B = 1$
writing it back to $(8)$: $(y + z) = 18 \tag{11}$
and writing this back to $(7)$
$72 \times 4 (18 - (y + z)) = 72 \times 4 \times 0 = 0$
which means the determinant of $(6)$ is zero and therefore can only have one solution:
$A = \frac{72}{2 \times 18} = 2$
using this in ${(5)}$: $x = 4$ and $yz = 72 \tag{12}$
using $(11)$ we get $y = 18 - z$, and substituting it to $(12)$ we get
$(18 - z)z = 72$
$z^2 - 18z + 72 = 0$
$z_{1,2} = \frac{18 \pm \sqrt{18^2 - 4 \times 72}}{2} = \frac{18 \pm \sqrt{18 \times 18 - 16 \times 18}}{2} = \frac{18 \pm \sqrt{36}}{2} = \frac{18 \pm 6}{2}$
$z_1 = 12 \Rightarrow y = 6$ per $(11)$
$z_2 = 6 \Rightarrow y = 12$ per $(11)$
so the 2 solutions for $z$ actually the same solution for the original statements, resulting in the triple (4, 6, 12).
Without loss of generality, suppose that $x\geq y\geq z$. From the given system of Diophantine equations, we obtain an Egyptian fraction problem: $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{yz+zx+xy}{xyz}=\frac{144}{288}=\frac12\,.\tag{*}$$ Since $xyz=288>0$, the number of variables with negative values among $x$, $y$, and $z$ is either $0$ or $2$. We consider two cases.
Case I: $x>0>y\geq z$. Let $u:=-y$ and $v:=-z$. Then, $$\frac{1}{x}-\frac1{u}-\frac1{v}=\frac{1}{2}\,.$$ Thus, $\dfrac{1}{x}>\dfrac12$, making $x<2$. Therefore, $x=1$. This implies $$yz=xyz=288$$ and $$y+z=x(y+z)=144-yz=144-288=-144\,.$$ Consequently, the polynomial $$q(t):=t^2+144t+288$$ has two roots $y$ and $z$. It is easily seen that $q(t)$ has no integer roots, so this case is invalid.
Case II: $x\geq y\geq z>0$. Then, $$\frac{3}{z}\geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac12\,.\tag{#}$$ This shows that $z\leq 6$. Furthermore, it is clear that $z>2$. Hence, there are four possible values of $z$, which are $3$, $4$, $5$, and $6$.
If $z=6$, then by (#), we conclude that $x=6$ and $y=6$. However, $xyz\neq 288$. This subcase yields no solutions.
If $z=5$, then this is impossible, as $xyz=288$ implies that $z$ divides $288$. This subcase is eliminated.
If $z=4$, then $$xy=\dfrac{288}{z}=\dfrac{288}{4}=72$$ and $$x+y=\dfrac{144-xy}{z}=\dfrac{144-72}{4}=18\,.$$ Thus, $t=x$ and $t=y$ are the roots of the quadratic polynomial $$t^2-18t+72=(t-6)(t-12)\,.$$ This means $x=12$ and $y=6$.
If $z=3$, then $$xy=\dfrac{288}{z}=\dfrac{288}{3}=96$$ and $$x+y=\dfrac{144-xy}{z}=\frac{144-96}{3}=16\,.$$ Thus, $t=x$ and $t=y$ are the roots of the quadratic polynomial $t^2-16t+96$, but this polynomial has no real roots.
In conclusion, all integer solutions $(x,y,z)$ to the required system of Diophantine equations are permutations of $(4,6,12)$.
Remark. Note that all $(x,y,z)\in\mathbb{Z}^3$ that satisfy (*) are permutations of the triples listed below. $$(1,-3,-6)\,,\,\,(1,-4,-4)\,,\,\,(k,2,-k)\,,\,\,(4,3,-12)\,,\,\,(5,3,-30)\,,$$ $$(6,6,6)\,,\,\,(10,5,5)\,,\,\,(20,5,4)\,,\,\,(12,6,4)\,,\,\,(8,8,4)\,,$$ $$(42,7,3)\,,\,\,(24,8,3)\,,\,\,(18,9,3)\,,\text{ and }(12,12,3)\,,$$ where $k$ is any positive integer.