Prove that,if $n$ is an integer, then $n^2 + 15n = 1$ is odd

Question

I'm proving this by cases and have so far:

Case 1:

Say $n$ is an even integer. Since $n$ is even, then $n=2m$ for some integer $m$. So $n^2-n+5 = (2m)^2-2m+5 = 4m^2-2m+5 = 2(2m^2-m)+5$....

Case 2:

Say $n$ is an odd integer. Since $n$ is odd, then $n=2x+1$ for some integer $x$. So $n^2-n+5 = (2x+1)^2-(2x+1)+5 = 4x^2+2x+5$....

How can I manipulate the equation in each case to show that is is odd?

Answer 1

If $n = 2m$ is even then you have $n^2 -n + 5 =2(2m^2 -m) + 5$. And that is $2(2m^2 - m) + 4 + 1 = 2(2m^2 - m +2) + 1$. Which is odd.

If $n = 2x + 1$ is odd then you have $n^2 - n + 5 = 4x^2 +2x + 5$. And that is $2(x^2 + x) + 5 = 2(x^2 +x) + 4 + 1 = 2(x^2 + x + 2) + 1$. Which is odd.

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It might be better to point out:

$odd*odd = odd$ and $odd*even = even$ and $even*even = even$ and $odd \pm odd = even$ and $even \pm even = even$ and $even \pm odd = odd$. (If you don't know that then prove that to yourself right now.)

Then

Case 1: $n$ is even. So $n^2$ is even. And $n$ is even. So $n^2 - n$ is even. And $5$ is odd. So $n^2 - n + 5$ is odd.

Case 2: $n$ is odd. So $n^2$ is odd. And $n$ is odd. So $n^2 - n$ is even. And $5$ is odd. So $n^2 - n + 5$ is odd.

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By the way. In your body you never actually said that you were trying to prove $n^2 - n + 5$ is odd.

And in your title you say you are trying to prove "$n^2 +15n =1$ is odd." This makes no sense. $n^2 + 15n =1$ is an equation. An equation is a sentence. A sentence is not a number. A sentence can not be odd. "prove $n^2 + 15n = 1$ is odd" makes as much sense as "Prove 'Dolphins eat fish' is gray". "Dolphins eat fish" can't be any color.

Also if $n$ is an integer then $n + 15$ is an integer. So if $n^2 +15n = n(n+15) = 1$ then the only pair of integers that multiply to $1$ is $1*1$ and $-1*-1$. So either $n = 1$ and $n+15 = 1$. Or $n = -1$ and $n+15 = -1$. Both of those are impossible.

ALSO.... if $n$ is even then $n^2$ is even and $15n$ is even and $n^2 + 15n$ is even. So $n^2 +15n = 1$ is not possible if $n$ is even because $1$ is odd. If $n$ is odd then $n^2$ is odd and $15n$ is odd and $n^2 +15n$ is even. So $n^2 + 15n = 1$ is not possible if $n$ is odd because $1$ is odd. So $n^2 + 15n = 1$ is impossible if $n$ is even or odd. So if $n$ is an integer then $n^2 + 15n \ne 1$. And if $n^2 + 15n = 1$ then $n$ is not an integer.

If you know the quadratic equation then

$n^2 + 15n = 1$ means $n^2 + 15n - 1= 0$ means $n = \frac {-15 \pm \sqrt{15^2 -4(-1)}}2 = \frac {-15 \pm \sqrt {229}}2$ where are not integers.

Answer 2

An alternative approach is by induction. Let $$P(n) : n^2-n+5 = 2m+1 \text { for some } m\in\mathbb{Z}.$$ $P(0)$ is true since $0^2-0+5=5$, which is odd.

Since we are considering both positive and negative integers we need to show that $P(n)\implies P(n+1)$ and $P(n)\implies P(n-1)$.

If $P(n)$ holds, we have \begin{align} (n+1)^2-(n+1)+5&=n^2+2n+1-n-1+5\\ &= (n^2-n+5)+2n, \end{align} which is an odd number plus an even number, hence it is odd.

Similarly, we have \begin{align} (n-1)^2-(n-1)+5&=n^2-2n+1-n+1+5\\ &=(n^2-n+5)-2n+2\\ &=(n^2-n+5)-2(n-1), \end{align} which is an odd number minus an even number,, hence also odd.

Therefore $n^2-n+5$ is odd for all $n\in\mathbb{Z}$.

Answer 3

This true for every polynomial with following form:

$(2k±1) x^n + (2k_1 ±1) x ^m + . . . +1$

If the number of terms is odd, because when the number of terms having factor x is even the sum of these terms will be even.This is clear when $x = 2p ±1$ is, in fact we have:

$x^ q=(2p ±1)^q= 2m ±1$

$c= 2k ±1$

$c \times x^q= 2t ±1$

That is any power of an odd number is odd, the results will also be odd if the coefficient c is odd. Now when the number of terms of odd terms is even the sum of them will also be even.It is far clear when x is even, the sum of terms is always even.

Here the number of terms is 3 and coefficients of x are 1 and 15.