Is $n^{2} + 5n + 1$ ever an even integer for a chosen natural number n?

Question

How is it shown that this number is odd for all natural numbers? Can someone show me why this is not an even integer using induction?

Thank you in advance.

Answer 1

If $n$ is even then $n^2$ is even $5n$ is even then $n^2+5n+1$ is odd.

If $n$ is odd then $n^2$ is odd $5n$ is odd then $n^2+5n+1$ is odd.

If you really need induction:

Hint:

$$(n+1)^2+5(n+1)+1=(n^2+5n+1)+(2n+6)$$

and once $n^2+5n+1$ is odd (by hypothesis) and $2n+6=2(n+3)$ is even then $(n+1)^2+5(n+1)+1$ is odd.

Answer 2

There are two cases :

• $n$ is even or $n=2k$

then $n^2+5n+1$

$=4k^2+10k+1$

$=2(2k^2+5k)+1$

$=2K+1 \implies n$ is odd.

• $n$ is odd or $n=2p+1$

then $n^2+5n+1$

$=(2p+1)^2+5(2p+1)+1$

$=4p^2+4p+1+10p+5+1=2(2p^2+7p+1)+1$

$=2K'+1 \implies n$ is odd.

in all cases, $n$ is odd.

Answer 3

Saying that $n^2+5n+1$ is even is equivalent to saying $n^2+5n$ is odd. We have $n^2+5n=n(n+5)$.

$(1)\qquad n$ even $\implies n+5$ odd.

$(2)\qquad n$ odd $\implies n+5$ even

An even number times an odd number is always even, so $n^2+5n$ is always even, and thus $n^2+5n+1$ is always odd.

Answer 4

For $ n=1$, we get $1^2+5\cdot 1+1=7$ so we have an odd number. Now, let's suppose that $ n^2+5n+1$ is odd. We have to prove that $(n+1)^2+5 (n+1)+1$ is odd. But $(n+1)^2+5(n+1)+1=n^2+2n+1+5n+5+1=(n^2+5n+1)+2(n+3) $ which is an odd number since by hyphotesis $ n^2+5n+1$ is odd and $2(n+3) $ is even. Hence $ n^2+5n+1$ is an odd number for every $ n\in\Bbb {N} $.

Answer 5

You don't need induction: just use Lil' Fermat and congruences: $$n^2+5n+1\equiv n+n+1=2n+1\equiv 1\mod 2.$$

If you absolutely want to use induction, here is the inductive step: $$(n+1)^2+5(n+1)+1-(n^2+5n+1)=2n+6=2(n+3), $$ hence, $(n+1)^2+5(n+1)+1$ and $n^2+5n+1$ have the same parity.

Answer 6

By Gauss' trick, $$ n^2+5n+1 = 2\times (3 + 4 + \cdots + (n+2)) +1 $$ According to an unverifiable tradition, Gauss discovered when he was ten years old that he could sum the terms of an arithmetic sequence by writing the sequence once from left to right, and once from right to left. By adding the two lines, one sees that the sum of the columns are all equal, in our case this sum is $(n+5)$. As there are $n$ columns in the sum, it follows that $2S=n\times (n+5)$ $$ \begin{array}{cccccc} S & = & 3 & +&4&+&\cdots &+&(n+1)&+&(n+2)\\ S & = & (n+2) & +&(n+1)&+&\cdots &+& 4&+&3\\ \hline 2S &=& (n+5)&+&(n+5)&+&\cdots&+&(n+5)&+&(n+5) \\ &=& n\times(n+5) \end{array} $$

Answer 7

so many ways.

But if you want induction:

1) $0^2 + 5*0 + 1 = 1$ is odd.

2) If $q$ is odd, then $q + 2k$ is odd.

If $n^2 + 5n + 1$ is odd then $(n+1)^2 + 5(n+1) + 1 =$ $ n^2 + 2n + 1 + 5n + 5 + 1 = $ $n^2 + 5n + 1 + 2n + 6 = $ $n^2 + 5n + 1 +2(n+3)$ is odd.

But there are are much easier ways. Such as:

$n^k \text{ is odd } \iff n \text { is odd }$ (assuming $k \ge 1$).

$even*n $ is always even.

$odd*n \text { is odd } \iff n \text{ is odd }$.

$odd + odd = even; odd + even = odd; even + even = even$

So... $n^2 + 5n + 1 = $ $odd/even + odd/even + odd$ $ = (odd + odd)/(even + even) + odd = even + odd$ $ = odd$.

Answer 8

If $n^2+5n+1$ is odd, then $(n+1)^2+5(n+1)+1=n^2+5n+1+2n+6$ has the same parity.

Answer 9

Hint: The parity of $n$ is different from the parity of $n+5$.

Can you take it from here?