jmc

Let $m^2$ and $n^2$ be the two-digit square numbers; we then have $4\le m,n\le 9$. Putting them next to each other yields a number $100m^2+n^2$, which must be equal to some other square $x^2$. Rearranging, we have $100m^2=x^2-n^2=(x+n)(x-n)$, so the RHS contains a factor of 100. The biggest possible square is 8181, whose square root is about 90.5, and the smallest is 1616, whose square root is about 40.2, so $41\le x\le 90$. To get the factor of 100, we have two cases:

1. Both $x+n$ and $x-n$ must be multiples of 5. In fact, this means $n=5$, $x$ is a multiple of 5, and $x-n$, $x$, and $x+n$ are consecutive multiples of 5. Trying possibilities up to $x=85$, we see that this case doesn't work.

2. One of $x+n$ and $x-n$ is a multiple of 25. Since $x+n=25$ is impossible, the simplest possibilities are $x-n=50$ and $x+n=50$. The case $x-n=25$ implies $x+n=4p^2$ for $(x+n)(x-n)$ to be a perfect square multiple of 100, and thus $57\le 4p^2\le 77$ from $41\le x\le 90$. The only possibility is $4p^2=64$, which leads to non-integral $x$ and $n$. The case $x+n=50$ requires $x-n=2p^2$ for $(x+n)(x-n)$ to be a perfect square. To have $x\ge 41$ we must have $x-n\ge 32$, and in fact the lower bound works: $(50)(32)=1600=40^2$. Thus $x=41$, and $x^2=\boxed{1681}$.