TITLE: Does another chain of three squares in this manner exist?
QUESTION [9 upvotes]: One of my computer science teachers came up with this neat little problem that he and I have both brute-forced and done some analytical work, but we haven't gotten far.
There exist chains of squares such that each square is the previous one but with a digit slapped on the end. For example...
$1\rightarrow 16 \rightarrow 169$
$1^2\rightarrow 4^2 \rightarrow 13^2$
$4 \rightarrow 49$
$2^2 \rightarrow 7^2$
$25 \rightarrow 256$
$5^2 \rightarrow 16^2$
There are an infinite* number of 2-square chains, but beyond the first one, we haven't found any 3-square chains and I have brute-forced it up to 20k (if I recall correctly) to look for more 3-square chains.
I have done some algebraic stuff, but I haven't come up with anything incredibly useful, unfortunately. Here's some of my work...
$n^2 = 10m^2 + k$
$n = m\sqrt{10} + \epsilon$
$n^2 = 10m^2 + 2\epsilon m\sqrt{10} + \epsilon^2$
$k = 2\epsilon m\sqrt{10} + \epsilon^2$
$0 = \epsilon^2 + 2\epsilon m\sqrt{10} + k$
Application of quadratic formula, simplifying, and noting that only the more positive answer is useful...
$\epsilon = -m\sqrt{10}+\sqrt{10m^2-k}$
While neat, this equation hasn't been of much use to me thus far... :P
*The continued fraction for $\sqrt{10}$ gives 2-square chains composed of the squares of the numerator and denominator.
So yeah...anyone have a way to construct 3-square chains? Even a proof that more than two exist would be very helpful.

REPLY [7 votes]: I believe your search is sufficient to prove there are no others.  If you let $m$ be the square root of the first in the chain and $n$ be the square root of the third, you have $100m^2+a=n^2$ with $9 \lt a \lt 100$.  This becomes $(n-10m)(n+10m)=a$, and $n \gt 10m$, so once $10m \gt 50$ there are no more solutions.  Another way to see it is that the smallest $n^2$ can be is $(10m+1)^2=100m^2+2m+1$ and with $m \gt 50$ this will not start with $m$ because of the carry.