Question 1199485
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Let the two perfect squares be {{{x^2}}} and {{{y^2}}}
{{{ x^2 - y^2 = 53 }}}

{{{ (x-y)(x+y) = 53 }}}

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  


{{{ 27^2 = 729 }}}
{{{ 26^2 = 676 }}}

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 {{{x^2 - y^2 = 63}}}
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).