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If m > 2, is √m an integer?

Statement :1 m = 2n, where n is a positive integer.
m = 2, 4 , 6 , 8 ..
Not sufficient

Statement :2 m − 1 is the product of two consecutive odd integers.

m-1=(2n-1)(2n+1) where n is an integer
m=4n^2

so, \(\sqrt{m}\) will be equal to 2n i.e an integer
Hence, sufficient

B is the answer
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If m > 2, is √m an integer?
or is \(m = n^2\) where \(n >=2\)

Statement (1) m = 2n, where n is a positive integer.
if \(m = 2* (2*2*2)\), then \(m = 4^2 \)and answer to D.S is Yes
if\( m = 2 * (3)\), then \(√m\) is not an integer and answer to D.S is No
Hence, insufficient.

Statement (2) m − 1 is the product of two consecutive odd integers.
Let the two consecutive odd integers be 2n-1 and 2n+1, where n is +ve because \(m>2\)
\(m - 1 = (2n-1) * (2n+1)\\
=> m - 1 = 4n^2 - 1\\
=> m = 4n^2\)
=> √m is an integer always.
Hence sufficient.
Answer: B
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If m > 2, is √m an integer?

(1) m = 2n, where n is a positive integer.

Let n=3 ,m=6 ,√m=not an integer
Let n=8,m=16, √m=4(integer)
Not sufficient.

(2) m − 1 is the product of two consecutive odd integers.
Let us take few example to check this out.
m-1= 1*3
m=4
√m = integer
m-1=3*5
m=16
√m=4
Sufficient
IMO B

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If m > 2, is √m an integer?

Stat1: m = 2n, where n is a positive integer.
at n= 1, √m= Not integer; n=8, √m= Integer. Not sufficient

Stat2: m − 1 is the product of two consecutive odd integers.
m-1= 3, m= 4; m-1= 15, m= 16; m-35=1, m= 36. So, √m= integer. Sufficient

So, I think B. :)
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