Prove by contradiction that for any even integer a and any odd integer b, 4 does not divide (a^2 + 2b^2).
Proposition: That 4k (k is any integer) = a^2 +2b^2, and a is even, and b is odd.
But 4k is even (product of any integer and 4), so a^2 must be even, as 2b^2 is even.
Dividing both sides by 4,
k=a^2/4 + 2b^2/4
but a is even, so a=2*n where n is an integer. a^2=4n^2
k= n^2 + b^2/2
But b is odd, so b^2/2 is not an integer.
Therefore, k cannot be an integer, so the proposition is contradicted.
To prove by contradiction that for any even integer a and any odd integer b, 4 does not divide (a^2 + 2b^2), we assume the opposite, which is that 4 does divide (a^2 + 2b^2).
So we start by assuming that there exists an integer k such that 4k = a^2 + 2b^2, where a is even and b is odd.
Since 4k is always even (as it is a multiple of 4), we can conclude that the sum of a^2 and 2b^2 must be even.
Now, let's consider the even number a^2. Since a is even, let's write a as 2n, where n is an integer. Substituting this into the equation, we get (2n)^2 = 4n^2.
So a^2 is always divisible by 4, making it even.
Now, let's consider the term 2b^2. Since b is odd, let's write b as 2m + 1, where m is an integer. Substituting this into the equation, we get 2(2m + 1)^2 = 2(4m^2 + 4m + 1) = 8m^2 + 8m + 2.
So 2b^2 is always divisible by 2, making it even.
Therefore, the sum of a^2 and 2b^2 is the sum of two even numbers, which is also even.
Now, dividing both sides of the equation 4k = a^2 + 2b^2 by 4, we get k = a^2/4 + 2b^2/4.
Since a is even, we can express it as a = 2n, where n is an integer. Substituting this, we get k = (2n)^2/4 + 2b^2/4 = 4n^2/4 + 2b^2/4 = n^2 + b^2/2.
But since b is odd, b^2/2 is not an integer (division by 2 introduces a fractional part).
Therefore, we have reached a contradiction. The assumption that 4 divides (a^2 + 2b^2) is false.
Hence, we can conclude that for any even integer a and any odd integer b, 4 does not divide (a^2 + 2b^2).