This question occurred to me and I thought it ought to be true; it holds if we replace 'perfect square' with 'prime number', for example. Of course it is not new and many others have asked it, see [1] for a example proof.
Looking for an informal approach, I enlisted the help of Gemini. Here's, not a proof, but a process, that I think always finds such a square.
As an example here is the number of the Golden Section, , to digits. We can ignore the decimal point because it doesn't make a material difference.
Take the square root
Take the first figures and ignore the decimal point, so we have the integer and also take the integer that is one larger, . Square both these and we should find one of the squares begins with the required digits.
This procedure should let you find a number whose square commences with your birth year, all you need is the calculator on your phone.
For example Queen Victoria was born in
[1] maths stack exchange 869383
