smc

Let the first term be $a$ and the common ratio be $r$, and WLOG let $r\ge 1$. The five terms are $a,ar,a{r}^2,a{r}^3,a{r}^4$, and the sum is $a(1+r+r^2+r^3+r^4)$. Clearly $r$ must be rational for all terms to be integers. If $r$ were an integer, it could not be $1$, since $a$ would equal $\frac{211}{1+r+r^2+r^3+r^4}$, which is not an integer. In fact, quickly testing $r=2,3,4$ shows that $r$ cannot be an integer. We now consider non-integers. If $r=\frac{x}{y}$ and $\gcd (x,y)=1$, then $a$ would have to be divisible by $y^4$, since $a{r}^4=a\frac{x^4}{y^4}$ is an integer. If $y=3$, then $a$ would have to be a multiple of $81$, which would make the five terms sum to at least $5\cdot 81$. It only gets worse if $y>3$. Thus, $y=2$, and $a$ is a multiple of $2^4=16$. Let $a=16k$. We now look at the last term, $a{r}^4=16k\frac{x^4}{16}=k{x}^4$. The smallest allowable values for $x$, given that $x$ cannot be even, are $3$ and $5$. If $x=5$, the last term will be way too big. Thus, $x=3$ is the only possibility. We now have a sum of $a(1+r+r^2+r^3+r^4)=211$, and we know that $r=\frac{3}{2}$ is the only possibility. The terms in the parentheses happen to equal $\frac{211}{16}$ when you plug them in, so $a=16$. Thus, the terms are $16,24,36,54,81$, and the first, third, and fifth terms are squares, with a sum of $133$, which is answer $\boxed{133}$.