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If x and y are positive integers such that x^2 – y^2 = 48, how many
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16 May 2021, 05:57
\(x^2 – y^2 = 48\) can be written as (x+y)*(x-y)=48
Since x and y are integers so even x+y and x-y will be integers
=> Product of two integers = 48
Possible values of the two integers can be
48*1, 24*2, 16*3, 12*4, 8*6 (as we have x+y * x-y so first number should be greater than the second)
So, we need to solve all of them and get the values of y
x+y = 48 and x-y = 1 => y=47/2 not an integer => not possible
x+y = 24 and x-y = 2 => y=22/2 => y = 11
x+y = 16 and x-y = 3 => y=13/2 not an integer => not possible
x+y = 12 and x-y = 4 => y=8/2 => y = 4
x+y = 8 and x-y = 6 => y=2/2 => y = 1
So, only 3 integer values of y are possible
So, answer will be B
Hope it helps!