# How many positive integers are there such that n^2+105 is a perfect square? A) 0 B) 2 C) 4 D) 6 E) infinitely many

embizze | High School Teacher | (Level 1) Educator Emeritus

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How many positive integers exist such that `n^2+105` is a perfect square:

Let `n^2+105=y^2` where `n,y in NN` (natural numbers or positive integers ** some include 0 in the natural numbers, others do not; here we do not include 0**)

Then `y^2-n^2=105`

So `(y+n)(y-n)=105`

105 factors as `1*105,3*35,5*21,7*15`

(1) y+n=105
y-n=1

2y=106 ==> y=53,n=52

(2) y+n=35
y-n=3

2y=38 ==> y=19,n=16

(3) y+n=21
y-n=5

2y=26 ==> y=13,n=8

(4) y+n=15
y-n=7

2y=22 ==> y=11,n=4

These are the only 4 ways to write 105 as a product of 2 integer factors.

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The answer is C. There are 4 positive integer n such that `n^2+105` is a perfect square -- namely n=4,8,16,and 52.

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