document.write( "Question 1199485: Prove that there is only one pair of (integer) perfect squares that differ by 53, and find it.\r
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Algebra.Com's Answer #833394 by math_helper(2461) $\"\"$ $\"About$

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document.write( "Let the two perfect squares be  and 
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document.write( "Factor 53 to find possible values.   Because 53 is prime, there are only two factors:  1 and 53.\r
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document.write( "x-y = 1
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document.write( "x+y = 53\r
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document.write( "Add both equations:
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document.write( "2x = 54
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document.write( "x = 27 \r
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document.write( "This implies y = 26  \r
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document.write( "Check:
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document.write( "729 - 676 = 53\r
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document.write( "Because we can only factor 53 into 1*53, there is only one pair of perfect squares that differ by 53.  \r
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document.write( "For example, if we try
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document.write( "x - y = 53
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document.write( "x + y = 1\r
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document.write( "Adding gives:
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document.write( "2x = 54   ==>  x = 27  and  y = -26\r
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document.write( "This also works, but the perfect squares are the same as before: 729 and 676
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document.write( "Therefore, there is only one pair of perfect squares (729 and 676) whose difference is 53. \r
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document.write( "----
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document.write( "  If you try a different difference, say 
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document.write( "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).\r
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