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Pick A Sequence Of Digits. Is There Always A Perfect Square Beginning With That Sequence?

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Edited by Richard Walker, Friday 25 September 2026 at 18:34

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, phi , to 10 digits. We can ignore the decimal point because it doesn't make a material difference.

16180339887

Take the square root

Square root of 16180339887 equals 40224.79320021422 times ellipsis

Take the first 10 figures and ignore the decimal point, so we have the integer 402249320 and also take the integer that is one larger, 402249321 . Square both these and we should find one of the squares begins with the required 10 digits.

4022479320 squared equals 16180339879827662400

4022479321 squared equals 16180339887872621041

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 1819

Square root of 1819 equals 42.649736224272246 times ellipsis
4264 squared equals 18181696
4265 squared equals 18190225

 

[1] maths stack exchange 869383

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