Is 4a^3+5a always a multiple of 3 for positive integers?

apple2357

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I am trying to work out if 4a^3+5a is always a multiple of 3 for positive integers.
I have thought about splitting a into odd and even numbers:
So if a is even (say 2k) we get 4(2k)^3 + 5(2k) = 32k^3 +10k = k(32k^2 +10) but this doesn't appear to go anywhere?

I have also thought about rewriting 4a^3+5a as perhaps 4a^3+ 4a + a = 4a(a^2+1) + a
so if a = even, a^2 is even etc.. but i can't conclude from that it is a multiple of 3.

Then i am thinking maybe it is not always a multiple of 3 ??

Any hints?
 
I am trying to work out if 4a^3+5a is always a multiple of 3 for positive integers.
I have thought about splitting a into odd and even numbers:
So if a is even (say 2k) we get 4(2k)^3 + 5(2k) = 32k^3 +10k = k(32k^2 +10) but this doesn't appear to go anywhere?

I have also thought about rewriting 4a^3+5a as perhaps 4a^3+ 4a + a = 4a(a^2+1) + a
so if a = even, a^2 is even etc.. but i can't conclude from that it is a multiple of 3.

Then i am thinking maybe it is not always a multiple of 3 ??

Any hints?

As for your "I am thinking" part, you'll need to prove it with just one example. Go!
 
So look for a counter example? I can't see one!

If you've tried enough values of a and didn't find a counterexample, then it's time to go back to looking for reasons why it might be true.

I would start by factoring it as a(4a^2 + 5). If a is itself a multiple of 3, then we're all set. If not, then a is either 3k+1 or 3k-1. See if each of those cases yields a multiple of 3.

I've convinced myself that it is true by another method that is more quirky and doesn't lend itself to hints.
 
I am trying to work out if 4a^3+5a is always a multiple of 3 for positive integers.
I have thought about splitting a into odd and even numbers:
So if a is even (say 2k) we get 4(2k)^3 + 5(2k) = 32k^3 +10k = k(32k^2 +10) but this doesn't appear to go anywhere?

I have also thought about rewriting 4a^3+5a as perhaps 4a^3+ 4a + a = 4a(a^2+1) + a
so if a = even, a^2 is even etc.. but i can't conclude from that it is a multiple of 3.

Then i am thinking maybe it is not always a multiple of 3 ??

Any hints?
With respect to 3, a number is either a multiple of 3, a multiple of 3 to 1 or a multiple of 3 + 2. In modular arithmetic notation where '=' means congruent to, we have for the factor a that a = 0 (mod 3), a = 1 (mod 3) or a = 2 (mod 3).

If a = 0(mod 3), then we are done as 4a^3+5a is a multiple of 3. Otherwise check if a = 1(mod 3) works by letting let a = 3r+1 and see if 4a2+5. If that does not work then check a = 2(mod 3) by letting a = 3r+2 and see if 4a2+5 is a multiple of 3. If none of these show that 4a^3+5a is always a multiple of 3 for positive integers, then it is not always a multiple of three. Using this method you are not required to show a counter example (although showing a counter example may be easier)
 
If you've tried enough values of a and didn't find a counterexample, then it's time to go back to looking for reasons why it might be true.

I would start by factoring it as a(4a^2 + 5). If a is itself a multiple of 3, then we're all set. If not, then a is either 3k+1 or 3k-1. See if each of those cases yields a multiple of 3.

I've convinced myself that it is true by another method that is more quirky and doesn't lend itself to hints.

Is this adequate:

a= 3k+1 , a(4a^2+5) = (3k+1)( 4(9k^2+6k+1) +5) = (3k+1) ( 36k^2+24k+9) = 3(3k+1)(12k^2+8k+3) , therefore always a multiple of 3
Same sort of thing for 3k-1

And i started off by splitting the integers into odd and even and didn't get anywhere!

By the way, whats your quirky alternative approach?
 
Last edited:
Is this adequate:

a= 3k+1 , a(4a^2+5) = (3k+1)( 4(9k^2+6k+1) +5) = (3k+1) ( 36k^2+24k+9) = 3(3k+1)(12k^2+8k+3) , therefore always a multiple of 3
Same sort of thing for 3k-1

And i started off by splitting the integers into odd and even and didn't get anywhere!

By the way, whats your quirky alternative approach?

First, what you did is what I had in mind, but can be done with a little less work. We just need to show that, if a is not a multiple of 3, then 4a^2 + 5 is a multiple of 3; so we just look at 4(3k+1)^2 + 5(3k+1) = (3k+1)(4(3k+1) + 5) = (3k+1)(12k+9) = 3(3k+1)(4k+3) and we have it (changing some signs for 3k-1).

What I did before was to write 4a^2 + 5 as 4a^2 - 1 + 6 in order to make a difference of squares, (2a+1)(2a-1) + 6. If a is not a multiple of 3, then neither is 2a, so either 2a-1 or 2a+1 is a multiple of 3. The result follows. The key idea here was to pull out a multiple of 3 and leave something that could be factored and proved to be a multiple of 3. I was also aware of the classic fact that the product of three consecutive integers is always a multiple of 3, because one of the three has to be.

What I suggested is a little longer but more sure to work.
 
And i started off by splitting the integers into odd and even and didn't get anywhere!
. Nicely done!
You need to understand what is really going on with odd and even integers so you do not wall into this trap again. Even numbers are multiples of two without a remainder, while odd numbers are multiples of two plus one (ex: 7=2*3+1, 21=2*20+1) Since you are dealing with multiples of 3 then use the fact that any number is a multiple of 3 or a multiple of 3 plus 1 or a multiple of 3 plus 2. Same for multiple of 4, etc.
 
I would prove this by induction...our hypothesis \(\displaystyle P_n\) can be:

\(\displaystyle 4n^3+5n=3k_n\)

We see the base case \(\displaystyle P_1\) is true, and so for our inductive step, we can add to the hypothesis:

\(\displaystyle \left(4(n+1)^3+5(n+1)\right)-\left(4n^3+5n\right)=3\left(4n^2+4n+3\right)\)

to get:

\(\displaystyle 4(n+1)^3+5(n+1)= 3\left(k_n+\left(4n^2+4n+3\right)\right)\)

If we define:

\(\displaystyle k_{n+1}=k_n+\left(4n^2+4n+3\right)\)

Then we have:

\(\displaystyle 4(n+1)^3+5(n+1)=3k_{n+1}\)

We have derive \(\displaystyle P_{n+1}\) from \(\displaystyle P_n\), thereby completing the proof by induction.
 
Each of the following are divisible by 3:


\(\displaystyle (a - 1)(a)(a + 1) \ = \)


\(\displaystyle a(a^2 - 1) \ =\)


\(\displaystyle a^3 - a\)


and \(\displaystyle \ \ 4(a^3 - a) \ = \)


\(\displaystyle 4a^3 - 4a\)


- - - - - - - - - - - - - - - - - - - - - - - - - -- --- - - - - - -



Rewrite the original expression \(\displaystyle \ \ 4a^3 + 5a \ \ \) as:


\(\displaystyle (4a^3 - 4a) \ + \ 9a\)


Each addend is divisible by 3.


Therefore, the original expression is _________________________ .
 
Each of the following are divisible by 3:


\(\displaystyle (a - 1)(a)(a + 1) \ = \)


\(\displaystyle a(a^2 - 1) \ =\)


\(\displaystyle a^3 - a\)


and \(\displaystyle \ \ 4(a^3 - a) \ = \)


\(\displaystyle 4a^3 - 4a\)


- - - - - - - - - - - - - - - - - - - - - - - - - -- --- - - - - - -



Rewrite the original expression \(\displaystyle \ \ 4a^3 + 5a \ \ \) as:


\(\displaystyle (4a^3 - 4a) \ + \ 9a\)


Each addend is divisible by 3.


Therefore, the original expression is _________________________ .
Very nicely done.
 
Each of the following are divisible by 3:


\(\displaystyle (a - 1)(a)(a + 1) \ = \)


\(\displaystyle a(a^2 - 1) \ =\)


\(\displaystyle a^3 - a\)


and \(\displaystyle \ \ 4(a^3 - a) \ = \)


\(\displaystyle 4a^3 - 4a\)


- - - - - - - - - - - - - - - - - - - - - - - - - -- --- - - - - - -



Rewrite the original expression \(\displaystyle \ \ 4a^3 + 5a \ \ \) as:


\(\displaystyle (4a^3 - 4a) \ + \ 9a\)


Each addend is divisible by 3.


Therefore, the original expression is _________________________ .

Lovely!
 
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