SOLUTION: Prove that there is only one pair of (integer) perfect squares that differ by 53, and find it. Thank you!

Let the two perfect squares be  and 









Factor 53 to find possible values.   Because 53 is prime, there are only two factors:  1 and 53.



x-y = 1

x+y = 53





Add both equations:

2x = 54

x = 27 



This implies y = 26  











Check:

729 - 676 = 53





Because we can only factor 53 into 1*53, there is only one pair of perfect squares that differ by 53.  



For example, if we try

x - y = 53

x + y = 1



Adding gives:

2x = 54   ==>  x = 27  and  y = -26



This also works, but the perfect squares are the same as before: 729 and 676

 

Therefore, there is only one pair of perfect squares (729 and 676) whose difference is 53. 



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  If you try a different difference, say 

you will find that there are three sets of perfect squares that have a difference of 63  (64 and 1,  144 and 81, 1024 and 961) and these correspond to the factorizations of 63  (7 * 9,   3 * 21, and 1 * 63,  respectively).