If $x, y, z$ are three distinct positive integers, where $x + y + z = 13$ and $xy, xz, yz$ form an arithmetic...

Question

If $x, y, z$ are three distinct positive integers, where $x + y + z = 13$ and $xy, xz, yz$ form an increasing arithmetic sequence, what is the value of $(x + y)^z$ ?

I've been trying to solve this by forming equations based off of formulas ($yz - xz = xz - xy$) and by trying to clean it up by systems of equations (I did $(13 - z)^2 = 13$), but I can't seem to find the solution.

Answer 1

If $xy$, $xz$ and $yz$ form an arithmetic progression, increasing in that order, then $z>y>x$ so in particular $y\leq5$ as otherwise $x+y+z\geq1+6+7=14$. This already leaves only $10$ triplets $(x,y,z)$ to check. We can continue to reduce the number of triplets by noting that $$(y-x)z=yz-xz=xz-xy=x(z-y),$$ which shows that $x$ divides $yz$ and $z$ divides $xy$. In particular, the latter shows that $$xy\geq z>y>x,$$ so $x>1$. Moreover, because $z=13-x-y$ this inequality shows that $$(x+1)(y+1)\geq14.$$ These conditions greatly reduce the possibilities for $(x,y,z)$; we are now left with $$(2,4,7),\qquad (2,5,6),\qquad (3,4,6).$$ We see that $z$ divides $xy$ only for the latter triplet, so $(x+y)^z=(3+4)^6=117\,649$.

Of course if the arithmetic progression is decreasing, then swapping $x$ and $z$ yields an increasing sequence and hence the solution found above. So the unique decreasing solution is $(x,y,z)=(6,4,3)$ with $(x+y)^z=1000$.

Answer 2

This is actually not too bad to brute force since $x + y + z = 13$ and they are all positive integers. There are actually only $11$ possible values for $x$: $1, 2, 3, \ldots, 11$.

However, with a bit of analysis, we can reduce the number of cases.

Since $xz - xy = yz - xz$, we have that $2xz = xy + yz = y(x+z) = y(13-y)$. Since \begin{align*} (x+z)^2 \ge 4xz = 2y(13-y), \end{align*} we have that \begin{align*} (13-y)^2 \ge 2y(13-y). \end{align*} Since $0 < y < 13$ (otherwise $x = z = 0$ which is impossible since $x$ and $z$ are positive integers), then $13 - y > 0$ so we can divide both sides by $13-y$ without changing the sign to get \begin{align*} 13-y \ge 2y, \end{align*} so this implies $y \le 4$. We have 4 cases to consider.

1. If $y = 1$, then $x + z = 12$ and $xz = 6$. No solution.
2. If $y = 2$, then $x + z = 11$ and $xz = 11$. No solution.
3. If $y = 3$, then $x + z = 10$ and $xz = 15$. No solution.
4. If $y = 4$, then $x + z = 9$ and $xz = 18$. This gives $(x,z) = (6,3)$ or $(3,6)$.

Hence, $(x+y)^z$ is either $(6+4)^3 = 1000$ or $(3+4)^6 = 117649$.