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x and y are a pair of positive integers such that both their sum and d

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x and y are a pair of positive integers such that both their sum and d  [#permalink]

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New post 17 Aug 2019, 19:46
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x and y are a pair of positive integers such that both their sum and difference equal a perfect square. If x+y<100, how many such pairs of integers are there?

A. 2
B. 4
C. 9
D. 12
E. 16
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x and y are a pair of positive integers such that both their sum and d  [#permalink]

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New post 17 Aug 2019, 20:32
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The possible perfect squares <100 are : 1,4,9,16,25,36,49,64,81
5 of them are odd, and 4 of them are even.

if x+y and x-y are equal to two perfect squares, then the two perfect squares must be either even or odd.
for example, if we assume that x+y = 9 (odd) and x-y = 1 (odd), then x = 5 and y = 4
but if we assume that x+y = 16 (even) and x-y = 1 (odd), there will be no valid integer values.

the number of valid pairs for the odd perfect squares = 5C2 = 10
the number of valid pairs for the even perfect squares = 4C2 = 6
total = 16
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Re: x and y are a pair of positive integers such that both their sum and d  [#permalink]

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New post 17 Aug 2019, 21:25
Let's start with value of positive squares below 100,

1,4,9,16,25,36,64,81..

given condition,
Sum and difference should be perfect square too.

Now observe that, for odd square there should be one even integer and one odd integer, (such that their sum or difference will be odd)

So every odd square will have sum or difference as odd. Similarly, even squares will have even integers or both odd integers such that sum/diff will be even.

Thus,

Starting with 1 = 0 pairs as there is no odd square less than 0 which can be the difference b/w two integers.

4 = again 0

9 = 1 pair as there is 1 odd square less than 9

Eg ( 5,4) = 5+4 = 9/5-4=1

16 = 1 pair as there is 1 even square less than 16

25 = 2 pairs (9,1)
Eg (13,12) ; (17,8)

36 = 2 pairs (16,4)

49 = 3 pairs ( 25,9,1)

64 = 3 pairs (36,16,4)

81 = 4 pairs ( 49,25,9,1)

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Re: x and y are a pair of positive integers such that both their sum and d  [#permalink]

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New post 18 Aug 2019, 10:29
Is there any easier way to get the pairs ?
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x and y are a pair of positive integers such that both their sum and d  [#permalink]

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New post 18 Aug 2019, 12:01
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ManjariMishra wrote:
Is there any easier way to get the pairs ?


hi Manjari,

let s=perfect square
start with two equations:
x+y=s1
x-y=s2
adding,
2x=s1+s2→
x=(s1+s2)/2
because x is integer, sum of squares must be even
so s1 and s2 must be either both even or both odd
there are 4 perfect even squares<100
so possible even pairs=(4*3)/2=6
there are 5 perfect odd squares<100
so possible odd pairs=(5*4)/2=10
6+10=16 total possible pairs

I hope this helps,
gracie
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x and y are a pair of positive integers such that both their sum and d   [#permalink] 18 Aug 2019, 12:01
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