# Prove algebraically that ##n^3+3n-1## is odd

• chwala
In summary, the conversation discusses different approaches to proving that the expression n^3+3n-1 is odd for all positive integers n. The first approach is to prove it using parity rules, showing that the expression is always odd regardless of whether n is odd or even. Another approach is to use induction, showing that the expression is odd for n=0 and then proving that the difference between f(n+1) and f(n) is always even. The discussion also touches on whether students would be penalized for using a different approach than what was expected, with the general consensus being that as long as the technique is correct, the result should not be penalized.

#### chwala

Gold Member
Homework Statement
Prove algebraically that ##n^3+3n-1## is odd for all positive integers ##n##.
Relevant Equations
Algebra
This is a past paper question; Find the solution here; Well understood

Now i was just thinking along this lines;

Let ##n=x##,
then ##f(x)=x^3+3x-1##
##f(x) =x(x^2+3)-1##

since ##x∈ℤ^{+}## then ##x(x^2+3)## will always be even implying that ##x(x^2+3)-1## is odd.
Would this approach hold or i have to stick with ms? Thanks.

Maarten Havinga
I prefer transformation of
$$n^3+3n-1=(n-1)^3+3n^2$$
Say n is odd (n-1)^3 is even and 3n^2 is odd so it is odd.
Say n is even (n-1)^3 is odd and 3n^2 is even so it is odd.
So we can say it is odd.

[EDIT]
$$n^3 \equiv n(mod\ 2)$$
$$n^3+3n-1 \equiv 4n -1 \equiv -1 \equiv 1 (\mod 2)$$ So it is odd.

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Why not use the parity rules O + O - O = O and E + E - O = O directly?

malawi_glenn said:
Why not use the parity rules O + O - O = O and E + E - O = O directly?
@malawi_glenn how to show this directly? I just looked at parity...some work though required...you still have to let ##n=2k+1## for odd numbers and let ##n=2k## for even numbers. Note that this was already used on the attached mark scheme guide.

...or you have a more direct way?

if n is odd, then you have
O2 + O*O - O = O

if n is even, then you have
E2+ O*E - O = O

malawi_glenn said:
Why not use the parity rules O + O - O = O and E + E - O = O directly?

chwala said:
@malawi_glenn how to show this directly? I just looked at parity...some work though required...you still have to let n=2k+1 for odd numbers and let n=2k for even numbers.
The first one above doesn't take much work.
##(2k+1) + (2m+1) - (2n+1) = 2(k + m + n + 1) + 1##, which is an odd number. The second one can be shown just as easily.

The basic ideas are that
• odd + odd = even
• odd + even = odd
• even + even = even
• odd * odd = odd
• even * even = even
• odd * even = even
You can manipulate these equations to get another set that involves subtraction and division.

chwala and malawi_glenn
@OP Don't confuse the reader with renaming to ##x## for no reason at all. Let ##n## be a natural number and give an argument why ##n^3+3n-1## is odd. For example ##n(n^2+1)+(2n-1)## is the sum of even and odd, because ##n(n^2+1)## is even for all ##n##.

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PeroK and chwala
nuuskur said:
Don't confuse the reader with renaming to x for no reason at all.
Which post are you referring to?

nuuskur
malawi_glenn said:
Which post are you referring to?
The very first one. We have ##n## and then it suddenly changes to ##x##, it's confusing.

malawi_glenn
Mark44 said:
The first one above doesn't take much work.
##(2k+1) + (2m+1) - (2n+1) = 2(k + m + n + 1) + 1##, which is an odd number. The second one can be shown just as easily.

The basic ideas are that
• odd + odd = even
• odd + even = odd
• even + even = even
• odd * odd = odd
• even * even = even
• odd * even = even
You can manipulate these equations to get another set that involves subtraction and division.
Yes, these rules are pretty handy to know.
@op: try to prove these. Then use them basically whenever you can

chwala
malawi_glenn said:
if n is odd, then you have
O2 + O*O - O = O

if n is even, then you have
E2+ O*E - O = O
That all fine, except that notation may not be part of the course. Moreover, the marking scheme suggests to me that an approach from first principles is expected.

chwala
PeroK said:
That all fine, except that notation may not be part of the course. Moreover, the marking scheme suggests to me that an approach from first principles is expected.
Would a student be penalised for using an approach other than what was expected? Like in this case using parity? Or it depends on examining body...

chwala said:
Homework Statement:: Prove algebraically that ##n^3+3n-1## is odd for all positive integers ##n##.
Relevant Equations:: Algebra

This is a past paper question; Find the solution here; Well understood

View attachment 318976

Now i was just thinking along this lines;

Let ##n=x##,
then ##f(x)=x^3+3x-1##
##f(x) =x(x^2+3)-1##

since ##x∈ℤ^{+}## then ##x(x^2+3)## will always be even implying that ##x(x^2+3)-1## is odd.
Would this approach hold or i have to stick with ms? Thanks.
If you were looking for an esoteric solution, try this:

Let ##f(n) = n^3 + 3n - 1##. Note that ##f(0) = -1## is odd. We will show by induction that ##f(n)## is odd for all integers ##n##.

Now$$f(n+1) - f(n) = (n+1)^3 + 3(n+1) - n^3 - 3n = 3(n^2 + n + 2)$$Let ##g(n) = n^2 + n + 2## and note that ##g(0) = 2## is even. Now:$$g(n+1) - g(n) = 2(n+1)$$The difference is even, hence by induction ##g(n)## is even for all integers ##n##. Hence ##f(n+1) - f(n)## is even for all integers ##n##. Hence ##f(n)## is odd for all integers ##n##.

Astronuc and chwala
PeroK said:
That all fine, except that notation may not be part of the course. Moreover, the marking scheme suggests to me that an approach from first principles is expected.
It's just schematic.
I do that notation in my math class (but swedish letters U for odd (udda) and J för even (jämn)) :-D

chwala said:
Would a student be penalised for using an approach other than what was expected? Like in this case using parity? Or it depends on examining body...
NO!!! If the technique is correct, then the correct result will be obtained.

If the problem statement doesn't specify what methods are allowed, then, in my opinion, it's in extremely bad taste to penalise students.

As an example, we do ##\varepsilon-\delta## proofs early in analysis tutorials to get comfortable with the definition. In test assignments we specifically state that prove the limit is so and so by definition of limit (there is no ambiguity here).

chwala
chwala said:
Would a student be penalised for using an approach other than what was expected?

nuuskur said:
NO!!! If the technique is correct, then the correct result will be obtained.
I disagree. If the correct result was obtained, but not by a specific technique that was explicitly stated, then the student almost certainly would be penalized.

I do agree, though, if there isn't a specific method required, then no penalty, which is what the following is saying.
nuuskur said:
If the problem statement doesn't specify what methods are allowed, then, in my opinion, it's in extremely bad taste to penalise students.

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chwala
chwala said:
This is a past paper question
Then the only consideration should be getting maximum possible marks. You can do this by using the methods that are taught in the syllabus to solve the problem, and the method that is taught for solving questions of the form "prove something is odd for all ## n \in \N ##" is to consider odd and even separately.

This appears to be from a public exam similar to a UK 'A'-level. Markers are paid about £3 per script for these, which leaves seconds to decide on each mark: they do this by comparing it to the marking scheme, so make it easy for them to give you full marks.

The instructions to markers will say something like
If a student uses a method which is not explicitly covered by the marking instructions the same principles of marking should be applied. Credit should be given to any valid methods. Examiners should seek advice from their senior examiner if in any doubt.
so if you do go "off-piste" you might be lucky, but if the marker is falling behind on their quota you might not. It is typically this kind of thing that gets picked up in a re-mark, but why go to the time, trouble, cost and risk of a re-mark when it is easier to use the method you were taught first time? There are no marks for being a smart-arse.

Disclaimer: I know what I am talking about: I used to be a real smart-arse.

chwala