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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
A. 2
B. 4
C. 9
D. 12
E. 16
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
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
General Discussion
17 Aug 2019, 21:25
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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)
Posted from my mobile device
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)
Posted from my mobile device
