Show that if $n^3-7n^2 + n$ is an odd number then n is an odd number.

Question

Need tips on how to think on solving the question:

Show that if $n^3-7n^2 + n$ is an odd number then $n$ is an odd number.

Answer 1

Hint. An odd number times an odd number is an odd number; an even number times either an even number or an odd number is an even number. So, if $n$ is odd, then

• Are $n^2$ and $n^3$ even or odd?
• What about $-7n^2$?
• What about $n^3-7n^2+n$ as a whole?

Then consider what happens if $n$ is even. Finally, consider that as a logical rule of inference,

$$\text{"If $n^3-7n^2+n$ is an odd number then $n$ is an odd number"}$$

is equivalent to

$$\text{"If $n$ is an even number then $n^3-7m^2+n$ is an even number"}$$

(that is to say, the contrapositive of a statement is logically equivalent to that statement). Can you put the pieces together into the desired demonstration?

Answer 2

Proof by cases:

Since $n$ is an integer, $n$ can be either even or odd.

If $n$ is even, then $(even)^3-7(even)^2+(even)$, which must produce an even result.

Since $n\ne even$ because $n^3-7n^2+n=odd$, $n$ must also be odd.

Thus, if $n^3-7n^2+n$ produces an odd number, $n$ is also an odd number.

Answer 3

Note that given:

$f(n) = n^3 - 7n^2 + n = n(n^2 - 7n + 1); \tag 1$

if $n$ is even, then $f(n)$ must be even since by (1), $n \mid f(n)$ and $2 \mid n$; therefore, $f(n)$ odd precludes the case $n$ even, so $n$ must then be odd.

So I guess the way to think about it is to observe, if possible, that the factorization of $f(n)$ given in (1) holds.

Answer 4

Proof of the Contrapositive:

Assume $n$ is even.

Show that $n^3 -7n^2 +n$ is even.

Since $n$ is even: $n=2k.$

$n^3-7n^2+n =$

$ (2k)^3-7(2k)^2 +(2k)=$

$2(2^2k^3 -7\cdot 2 k^2 +k)$ , hence even.

Answer 5

Hint: $$n^3-7n^2+n=n(n-1)^2-5n(n-1)-5n.$$ Note that the first two terms are even.