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AyushAgrawal1005
ANS - C
(n+m)(n-m) = 105
105 = 105*1 (m=52 n=53)
105 = 35 *3 (m=1 n =19)
105 = 21*5 (m=8 n=13)
105 = 15 * 7 (m=4 n=11)

AyushAgrawal1005

I think since you have written 4 ordered pairs so answer option should be B instead of C. Is that correct? :)
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Yes, the answer should be B. Have edited it in the post. Thanks
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(m+n)(m-n)=105
Now 105=3×5×7
Therefore 105 has 8 factors so 4 pairs are possible.
Option B is the answer.

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Bunuel
How many ordered pairs (m, n) of positive integers satisfy the equation \(m^2 + 105 = n^2\) ?

A. 3
B. 4
C. 5
D. 6
E. 7

Are You Up For the Challenge: 700 Level Questions

Simplifying, we have:

105 = n^2 - m^2

105 = (n - m)(n + m)

Since 105 = 3 x 5 x 7, 105 can be expressed as a product of two positive integers, as follows:

1 x 105, 3 x 35, 5 x 21 and 7 x 15

For the pair of integers in each of the 4 multiplications above, we can solve uniquely for n and m. For example, if we set n - m = 1 and n + m = 105, we can uniquely determine that n = 53 and m = 52. Therefore, since there are 4 multiplications, there will be 4 solutions.

Answer: B
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Bunuel
How many ordered pairs (m, n) of positive integers satisfy the equation \(m^2 + 105 = n^2\) ?

A. 3
B. 4
C. 5
D. 6
E. 7


Are You Up For the Challenge: 700 Level Questions
Asked: How many ordered pairs (m, n) of positive integers satisfy the equation \(m^2 + 105 = n^2\) ?

m^2 - n^2 = 105 = 3*5*7
(m+n)(m-n) = 1*3*5*7

m + n = 105; m - n = 1; m = 53; n = 52; 1 solution
m + n = 35; m - n = 3; m = 19; n = 16; 1 solution
m + n = 21; m - n = 5; m = 13; n = 8; 1 solution
m + n = 15; m - n = 7; m = 11; n = 4; 1 solution
m + n = 7; m - n = 15; m = 11; n = -4; Not a solution

Only 4 ordered Pairs of positive integers are possible

IMO B­
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