ildude02 wrote:All statement says is m ^ 3 + m = 4a; => m(m ^2 +1) = 4a; where a is an integer.
Good work here- and you're right that you won't find an odd value for m that makes this work. We can prove this abstractly if we want to:
m(m^2 + 1) is divisible by 4
Notice that only one of the factors can be even. Either:
*) m is divisible by 4 (and m is even), or
*) m^2 + 1 is divisible by 4 (and m is odd).
The only question is: is it at all possible for m^2 + 1 to be divisible by 4?
The 'standard' way to do this is to note that since m is odd, m = 2k + 1, for some integer k:
m^2 + 1 = (2k + 1)^2 + 1 = 4k^2 + 4k + 2
Since 4k^2 + 4k is divisible by 4, when we add 2 we definitely will not have a multiple of 4. That is, there's no way for m^2 + 1 to be divisible by 4. The only possibility is that m is divisible by 4: Statement 1 guarantees that m is even, and it is sufficient. Statement 2 is also sufficient; the correct answer should be D, not B.
There's a more interesting way to see that m^2 + 1 cannot be divisible by 4. Notice that:
m^2 + 1 = (m^2 - 1) + 2 = (m-1)(m+1) + 2
Now, if m is odd, (m-1)(m+1) is the product of two consecutive even integers, and the product of two consecutive even integers must be divisible by 8 (one of them must be divisible by 4, and the other is even and will not be divisible by 4). So m^2 - 1 is always divisible by 8 if m is odd (try it with numbers if you want to be convinced!), and m^2 + 1 is always two larger than a multiple of 8 (and, of course, of 4)- and thus cannot be divisible by 8, or by 4.
Incidentally, where's the question from?
