M is positive integer, is M odd?

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by DanaJ » Sun Apr 19, 2009 1:06 pm
1. 2m^3 + 2m = 2(m^3 + m). Now, this number is divisible by 8, which means that, since 8 = 2*4, m^3 + m must be divisible by 4. In this case, you get that:

m^3 + m = m (m^2 + 1). Since this number is divisible by 4, it must also be an even number.

Remember this rule:

A number and its square are odd or even at the same time

This means that, if say a number x is odd, then x^2 will also be odd (this is quite logical, since x^2 = x*x and x is odd).

Now, coming back: m and m^2 are both odd or even at the same time. However, m^2 + 1 is the "opposite" of m: if m is odd, then m^2 + 1 is even; if m is even, then m^2 + 1 is odd. This means that no matter what, m(m^2 + 1) will always be even, since one of its elements will always be even. This is why 1 is not sufficient: it doesn't matter if m is even or odd, the product discussed will always be even.

2. is a bit less complicated. Since m + 10 is divisible by 10, this means that m is also divisible by 10. This means that m is indeed even, since 10 only evenly divides even numbers.

This is because of the general rule:

If a is divisible by b and c is divisible by b, then a - c will also be divisible by b
So 2 is sufficient.