# Integer expression

#### Ed Aboud

1. Homework Statement

Prove that the expression $$\frac{a(a^2 + a)}{3}$$ is an integer for all integers $$\geq 1$$

2. Homework Equations

3. The Attempt at a Solution

$$a(a^2 + a ) = 3q + r$$

r can be:

$$r = 0,1,2$$

for r = 0

$$\frac{a(a^2 + a)}{3} = q$$

q is an integer by the division algorithm.

When I try this attempt with 3q+1 and 3q+2 I can't get it to work.

1. Homework Statement

2. Homework Equations

3. The Attempt at a Solution

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#### Tedjn

This is not true for all positive integers. For example, setting a = 1 gives 2/3.

#### Ed Aboud

Sorry there is a correction $$\frac{a^3 -a}{3}$$

#### gabbagabbahey

Homework Helper
Gold Member
Sorry there is a correction $$\frac{a^3 -a}{3}$$
You might want to factor $a^3 -a$, and show that for all integers $a\geq1$, one of the factors will be divisible by 3.

#### Hurkyl

Staff Emeritus
Gold Member
3. The Attempt at a Solution

$$a^3 - a = 3q + r$$

r can be:

$$r = 0,1,2$$

for r = 0

$$\frac{a^3 - a}{3} = q$$

q is an integer by the division algorithm.
Where did q and r come from? Whenever you introduce new variables, you need to say what they are, and why they should have the properties you say they have. You need to say something like "By the division algorithm, we can select integers q and r such that a^3 - a = 3q + r, where 0 <= r < 3". If you don't explain the introduction of new variables, then your work is just nonsense.

Yes, I was able to figure out what you meant, but that's irrelevant; it would be completely analogous to saying "me food want" in your English classes.

When I try this attempt with 3q+1 and 3q+2 I can't get it to work.
Of course not. If r=1 or r=2, then (a^3-a)/3 is not an integer. That's what 'remainder' means, right? And r is supposed to be a remainder?

#### Count Iblis

Fermat's little theorem: a^2 Mod 3 = 1 if a is not a multiple of 3. So we always have that a^3 = a Mod 3.

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