Question

How many pairs of natural numbers are there the difference of whose squares is 39?

Answer 1

Hi there!
Here's the answer:

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Let the two consecutive No.s be x and y

Given Difference of squares of these No.s = 39

$(x^{2} - y^{2}) = 39$

= $(x+y)(x-y) = 39$


The factors of 39 are 1 , 3, 13, 39

39 = 39×1 = (x+y)(x-y)

Here, x+y = 39 and x-y = 1
solving gives 2x = 38
x = 19
y= 39 - x = 39 - 19 = 10
So this case is possible

39 = 13×3 = (x+y)(x-y)

Here, x+y = 13 and x-y = 3
solving gives 2x = 36
x = 8
y= 13 - x = 13 - 8 = 5
So this case is possible


Hence the No.s are (8, 5) and (20, 19)

•°• No. of pairs possible = 2


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