Find all integers $x$ that are at most $100000$ so that $S(11x) = x$. 
For a positive integer $x,$ let $S(x)$ denote the sum of the squares of the digits of $x$.


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*Find all integers $x$ that are at most $100000$ so that $S(11x) = x$.

*Prove or disprove whether there are infinitely many $x$ so that $S(11x) = x$.

I know that the two digit integers $x=50,73$ both work. Among the one digit numbers $x$, $S(11x) > x$. I think it's useful to observe that once $S(11x)$ exceeds $x$, for any $y$ larger than $x$ with only the units digit differing from $x, S(11y) > y,$ though I'm not sure how to prove this. Similarly, for any $y$ smaller than $x$ with only the units digit differing from $x, S(11y) < y$ if $S(11x) < x$. For the proof, I think one could start by writing $x = 100 a + 10b + c, y = 100a + 10b + d,$ where $a$ is a nonnegative integer and $b,c,d$ are digits. From this, one has $11x = 1100 a + 110 b + 11c, 11y = 1100a + 110b + 11d.$
 A: Your conjecture is reasonable but fails for $n=90$, as $S(11\cdot90) = 162>90$ but $S(11\cdot91) = 2<91$.
Let's fix $n \ge 1$ and find a bound on $S(11x)$ where $x$ has exactly $n$
digits:
$$ \overbrace{110...0}^{n+1} \le 11x \lt \overbrace{1100...0}^{n+2}, $$
which implies $S(11x) \le S(\overbrace{99...99}^{n+1}) = 9^2(n+1)=81n+81$. For
$n \ge 4,$ it can be shown by induction on $n$ that $81n+81 < 10^{n-1} \le x$
for any $x$ with $n$ digits. Thus there are no solutions with more than 3
digits, and the second question has a negative answer.
For $n=1$ there are no solutions by inspection.
For $n=2,3$ there are no more solutions besides $x=50,\ 73$. As a first bound,
note that if $x = S(11x),$ then by the previous argument,
$$x = S(11x) < 81\cdot4 = 324.$$
For $182 \le x \le 324,\ \ 3000 \le 11x \le 3564 \implies x = S(11x) \le S(3599) = 196.$ So any answer is smaller than 196.
A computer search shows the only 2 solutions are the ones you gave; however, for $x$ below $179$ it gets harder as $S(11 \cdot 179)>179$. I don't know if there is a fast manual way to find the answer.
