I've tried solving the same problem yesterday. I managed to brute-force my way to finding all fair and square (palindromes whose square root is a palindrome) numbers from 1 to $10^{14}$:
1, 4, 9, 121, 484, 10201, 12321, 14641, 40804, 44944, 1002001,
1234321, 4008004, 100020001, 102030201, 104060401, 121242121,
123454321, 125686521, 400080004, 404090404, 10000200001, 10221412201,
12102420121, 12345654321, 40000800004, 1000002000001, 1002003002001,
1004006004001, 1020304030201, 1022325232201, 1024348434201,
1210024200121, 1212225222121, 1214428244121, 1232346432321,
1234567654321, 4000008000004, 4004009004004
My solution was correct, and the second dataset was solved. But I couldn't find a proper way to calculate all fair and square numbers up to $10^{100}$.
I showed this to my wife this morning, and she noticed an interesting pattern of numbers within my list:
121, 10201, 1002001, 102030201, 10000200001, 1000002000001
484, 40804, 4008004, 400080004, 40000800004, 4000008000004
12321, 1002003002001,
Some fair and square numbers re-appear with space padding. Let's try beyond $10^{14}$. Adding some zeros to $1020302030406040302030201$, whose square root is $1010100010101$ - a palindrome!
Wish I had my wife with me when I solved this yesterday.
I don't have a mathematical explanation for this phenomena, but I guess that for some reason, every fair and square number beyond a certain boundary can be built by adding zeros to a smaller palindrome.
Best Answer
You were not able to prove it because it's not true. The OEIS has the first 17 of them as sequence A060792, and it is not known whether any more exist.