prove that there is only one pair of (integer) perfect squares that differ by 53, and find it

hi1234 Jan 18, 2024

#1**+2 **

*prove that there is only one pair of (integer) perfect squares that differ by 53, and find it*

Let's call the squares x^{2} and y^{2} so y^{2} – x^{2} = 53

Well, I couldn't figure out how to do it, so

I went to Desmos and plotted the curve of y = sqrt(x^{2} + 53)

and by careful scrutinization, discovered

the place where the curve crossed two

whole numbers on the grid. It wasn't

as hard as describing it sounds.

The answer is when x = 26 & y = 27 27^{2} – 26^{2} = 53

**729** – **676** = 53

I can't prove that's the only pair. Sorry.

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Bosco Jan 18, 2024