given that n is an integer, prove that n is odd if and only if 3n^-4n+1 is even (note ^ is square)?

Answer 1

if n is even then 3n^2-4n being even 3n^2-4n + 1 is even

if n odd then 3n^2-4n +1 is even
if 3n^2-4n + 1 is even then 3n^2+1 is even so 3n^2 is odd so n^2 or n is odd

proved

Answer 2

To prove if and only if statements, we have to prove necessity and sufficiency.

Given n is an integer, prove that n is odd if and only if
3n^2 - 4n + 1 is even.

Necessity. " → "

Assume that n is odd. Then, n can be expressed in the form
n = 2k + 1, for an integer k.

That means

3n^2 - 4n + 1 = 3(2k + 1)^2 - 4(2k + 1) + 1
= 3(4k^2 + 4k + 1) - 8k - 4 + 1
= 12k^2 + 12k + 3 - 8k - 3
= 12k^2 + 4k
= 2(6k^2 + 2k)

This shows that 3n^2 - 4n + 1 is even, since it can be expressed as 2 times something.

Sufficiency: "←"

Assume 3n^2 - 4n + 1 is even. Then 3n^2 - 4n + 1 can be expressed as 2k for some integer k. That is,

3n^2 - 4n + 1 = 2k

Subtract 1 both sides,

3n^2 - 4n = 2k - 1
n(3n - 4) = 2k - 1

This implies that the product of n(3n - 4) is odd.
Claim: n is odd.
Proof by contradiction; if n was not odd, then n would be even, and n(3n - 4) would be even. This contradicts the given statement that n(3n - 4) is odd.

Therefore n is odd.

Since we've proven necessity and sufficiency, it follows that
n is odd if and only if 3n^2 - 4n + 1 is even.

Answer 3

Since this is a quadratic equation, there exist two solutions for n.

If 3n^ - 4n +1 = even number, then 3n^ - 4n = odd number.

Splitting up 3n^ - 4n into n(3n - 4), the solutions are that the right half (n) and the left half (3n - 4) are both odd numbers. This proves that one of the solutions of n is an odd number.

For the other solution, 3n - 4 = odd. Since an odd number plus an even number is always an odd number, 3n = odd. Since it is stipulated that n is an integer, the odd number set to be equal to 3n has to be divisible by 3. Since any odd number divisible by 3, when divided by 3, yields an odd number, the other solution of n is odd as well.

Answer 4

3n^2 - 4n + 1 =

= n(3n - 4) + 1

An even number multiplied by any integer is even, so

for n=even, n(3n-4) is even. Add 1 and it is always odd.

The product of two odd numbers is always odd, so for n=odd

3n-4 is odd, so also n(3n-4) is always odd.

Add 1 and it is always even

Answer 5

hi

3n^2 - 4 n + 1
= (3 n - 1) ( n -1)
= (2n + (n - 1)) (n - 1)

Product of two number with same parity !!!

n even --> n - 1 odd --> product odd
n odd --> n - 1 even --> product even


bye