Possible values of $x+y+z$

Question

Given that $xyz=11(x+y+z)$, find the possible values of $x+y+z$ such that $x\leq y$ satisifies $x^2+y^2=2018$

I know that $x,y$, or $z$ should be divisible by $11$. I am having trouble going to the next step

Answer 1

Since $2018=2\cdot 1009$, we have only the divisors $1,2,1009,2018$, so we have $$ r_2(2018)=4(d_1(n)-d_3(n))=4, $$ where $$d_i(n)=\sum_{d\mid n,d\equiv i \bmod 4}1.$$ So there are only $4$ possibilities, up to sign and permutation, namely essentially only $$ 13^2+43^2=2018. $$ Then we solve $xyz=11(x+y+z)$ for all $4$ possibilities $(x,y)=(\pm 13,\pm 43)$. This is not solvable for integer $z$.

Answer 2

I think the problem also states that $x,y,z$ must be integers.

Assume that $x$ is divisible by $11$, then $x=11k$ ($k$ is an integer) and $(11k)^2+y^2=2018$.

We will have $y^2=2018-121k^2$. Because $y^2\ge 0$, we can prove that $k^2$ must be in this set of numbers: $(1;4;9;16)$.

We have four cases:

• $k^2=1\Rightarrow y=43.55$, result eliminated

• $k^2=4\Rightarrow y=39.17$, result eliminated

• $k^2=9\Rightarrow y=30.48$, result eliminated

• $k^2=16\Rightarrow y=9.05$, result eliminated

So there are no cases available for "$x$ is divisible by $11$" that satisfy $x^2+y^2=2018$, this means "$y$ is divisible by $11$" cannot be true as well.

Because of this, someone should attempt it by solving the equation $x^2+y^2=2018$ for integers $x,y$ instead, you can make a table that consists of $45$ cases from $1$ to $45$ for $x$ to find $y$ (or fewer cases if you use the condition $x\le y$ to eliminate some cases). This method is really tedious, but you should get $x=13$ and $y=43$, then you can solve for $z$.

However, $z=1.12$ which means $z$ is not an integer, this means there are no three integers $x,y,z$ satisfy all the conditions.

Answer 3

For real numbers, we have $$z=\frac{11x+11y}{xy-11}=\frac{22(x+y)}{(x+y)^2-2040}$$ subject to $-2\sqrt{1009}\leq x+y\leq 2\sqrt{1009}$

Answer 4

To use as little machinery as possible but at the cost of a bit of computation, note that we must have $\frac 12\cdot 2018 \lt y^2 \lt 2018$, which gives $33 \le y \le 44$, only $12$ possibilities. Trying each to check if $x$ is an integer gets us that the only choice for $x,y$ is $x=13, y=43$. That would require $13\cdot 43z=11(13+43+z)$ which clearly has no solution with $z \ge 1$ because the left side is much too big.

Answer 5

First $x^2+y^2=2018$ has as only solutions $(x,y)=(\pm13,\pm43),(\pm43,\pm13)$ so by the restriction $x\le y$ we have two solutions $(x,y)=(\pm13,43)$. This is not compatible with the equation $xyz=11(x+y+z)$ because in both cases we would have $$-z=\frac{330}{570}\\z=\frac{616}{548}$$ Thus there are no solutions.