write a convincing argument that if n is an integer and n^2 is even?

Question

then n is even

Answer 1

If I multiply an odd number by itself, I get an odd number.
If I multiply an even number by itself, I get an even number.
A square is a number multiplied by itself.
If the square is even, it cannot be an odd number multiplied by itself.
Therefore it's a even number mutilplied by itself.

Answer 2

If n^2 is even then it is a factor of 2 in the result. In order to take the square root of this result there must be another 2 also in the factor giving n a factor of 2 and therefore n is even.

Answer 3

If a number is odd it's in the form 2k+1 and if it's even it's in the form 2k for some integer k.

(2k)^2=4k^2 which is divisible by 2 so it's even

(2k+1)^2=4k^2+4k+1 so it's 2(2k^2+2k) +1 which is odd

So this shows the correspondence that you need. In other words, by contrapositive, if n is not even (it's odd) then n^2 is not even (it's odd).

Answer 4

n and n^2 contain the same prime factors but n^2 contains the factors raised to an even exponent. So if n^2 is eves it must contain the prime factor 2 raised to an even ex ponent,at least 2.
So n contained prime factor 2 and is even

Answer 5

there is none....an even integer squared will give you an even solution, and an odd integer squared will give you an odd solution
any value of n such that n is not odd will produce a positive even solution when squared, that includes negative numbers

Answer 6

Odd multiplied by the same Odd number will always be Odd.
Even multiplied by the same Even number will always be even.
Given the hypothesis, n is even.

Answer 7

An even number times an even number always gives back an even number

QFD