how to prove this, if n is an integer and n^2 is even then n is even?

Answer 1

This relies on unique decomposition of integers as products of prime powers.

Suppose n = p1^a1 . p2^a2 ... pk^ak
for some primes p1 < p2 < ... < pk and some a1, ..., ak ∈ Z+.
Then n^2 = p1^(2a1) . p2^(2a2) ... pk^(2ak).
If n^2 is even then we must have p1 = 2, and so 2 divides n also. So n is even.

We can also prove this with modular arithmetic:
Let n be any integer. Either n is congruent to 0 (mod 2) or n is congruent to 1 (mod 2).
If n is congruent to 1 (mod 2) then n^2 is congruent to 1.1 = 1 (mod 2). So n^2 is odd.
Thus is n^2 is even n must be congruent to 0 (mod 2) and hence n is even.

Another way of expressing this last argument is to say that any integer can be written as n = 2k + a where k is an integer and a is either 0 or 1. Then n^2 = 4k^2 + 2ak + a^2 = 2(2k^2 + ak) + a^2 will be divisible by 2 if and only if a^2 is divisible by 2. But a^2 = 0 for a=0, 1 for a=1. So n^2 will be even if and only if a = 0, i.e. n = 2k, i.e. n is even.

Answer 2

One way to prove this is by contradiction.

Let us assume that n is NOT even. (We want to prove that this CANNOT happen.) Remember that n is an integer and n^2 is even, as given by the problem.

Then, if n is not even, n is odd, and can be expressed in the form (2k + 1). That is,

n = 2k + 1, for some integer k.

It would then follow that

n^2 = (2k + 1)^2

Expanding that, we have

n^2 = (2k + 1)^2 = 4k^2 + 4k + 1
Factoring out 2 from the first two terms, we have

n^2 = 2(2k^2 + 2k) + 1

This implies that n^2 is odd (since it is of the form 2*[integer] + 1). However, this contradicts our supposition that n^2 is even, and this is a contradiction.

Therefore, it must be the case that n is even.

Answer 3

If n^2 is even then it has the prime factor 2 in its prime decomposition, i.e,
n^2 = (2^a1).(3^a2).(5^a3).....(p^ai)...

where i ranges as the number of the prime.(i.e. 1st prime-2, 2nd prime-3, etc.)
but since n^2 is the square of an integer n, every prime factor that made up n is squared to make up n^2.
Then it would appear that the square root of two appears as a factor of n, but that can't be, because V2 is not (by observation) an integer.
The only other explanation is that n^2 has a factor 2^(2a) where a can be any positive integer. This is the only way n can be an integer as well. Then n^2 is even. Taking the square root 2^(2a) in the factorisation becomes 2^a. Hence it is proven that n is even.
QED

Answer 4

n and n^2 have the same prime factors only the exponents of the primes of n^2 are twice the ones of n ( so they are even)
If n^2 is even 2 has an even exponent( at least 2) So n contains the factor 2 ==> is even

Answer 5

n^2 is even.
assume n is odd so n= 2k+1
n^2 = (2k+1)(2k+1)
n^2= 4k^2 + 4k +1
n^2 = 4k (k+1) +1
4k is even so 4k(k+1) is even --->
4k (k+1) +1 is odd
n^2= 4k (k+1) +1 is odd and it's inconsistent with what we supposed first (n^2 is even)

so n^2 is even ---> n is even

Answer 6

Well I don't really know how to PROVE it, but any odd number squared is odd and any even number squared is even. . .

sorry I guess I really cant help.

Answer 7

If two even numbers are multiplied together, it's even.

If two odd numbers are multiplied together, it's odd.

Therefore, if n^2 is even, n is even. It cannot be odd.

Answer 8

If two even numbers are multiplied together, it's even.
If two odd numbers are multiplied together, it's odd.
Therefore, if n^2 is even, n is even. It cannot be odd.

Answer 9

if x is even than x=2k, assume that 2k are sqrt able to integer so sqrt 2k=2k*.
n^2 is even than n^2=2k with k is integer.
n^2=2k
n=sqrt 2k because sqrt 2k=2k*. than n=2k* so n is even