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Positive integer n is equal to the difference of the squares

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Farina
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Positive integer n is equal to the difference of the squares [#permalink] New post   25 Jun 2020, 23:31
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Positive integer n is equal to the difference of the squares of x and y, where x and y are integers. If n is equal to 21, then which of the following could be the sum of x and y?
A. 0
B. 11
C. -1
D. -11
E. 121
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Re: Positive integer n is equal to the difference of the squares [#permalink] New post   26 Jun 2020, 07:57
Expert Reply
(x^2-y^2) = 21
=> (x+y)(x-y) = 21
Since x and y are integers
we can have following possibilities

7*3

3*7

21*1

1*21

-1*-21

-21*-1


out of the given options (-1) is the only value that fits into the solution.
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georgekaterji
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Re: Positive integer n is equal to the difference of the squares [#permalink] New post   26 Jun 2020, 08:46
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Farina wrote:
Positive integer n is equal to the difference of the squares of x and y, where x and y are integers. If n is equal to 21, then which of the following could be the sum of x and y?
A. 0
B. 11
C. -1
D. -11
E. 121


n= x2 - y2 (or y2-x2 but it wont make a difference in this case)

x2-y2 = (x-y)(x+y)

21= (x-y)(x+y)

x-y and x+y must be integers (since x and y are integers)

therefore,
21/(x+y) must produce an integer

only possible value for x+y to produce an integer from the given choices is: x+y=-1
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Re: Positive integer n is equal to the difference of the squares [#permalink] New post   14 Aug 2023, 04:54
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Re: Positive integer n is equal to the difference of the squares [#permalink]
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