# Prove that a perfect square is a multiple of some number?

1. Mar 27, 2013

### John112

How do you prove that: Every perfect square is either a multiple of 3 or one more than a multiple of 3?

There is something very basic that I don't seem to understand in the question.

Here is proof that someone gave me:

Every natural integer p must fall in one of the following case:
♠ p is a multiple of 3 and can be expressed as p=3n
♣ p is 1 more than a multiple of 3 and can be expressed as p=3n+1
♥ p is 1 less than a multiple of 3 and can be expressed as p=3n–1

Then obviously:

♠ If p=3n:
p² = (3n)² = 9n²
and p is naturally a multiple of 3.

♣ If p=3n+1:
p² = (3n + 1)² = 9n² + 6n + 1 = 3(3n² + 2n) + 1
and p² is logically 1 more than a multiple of 3.

♥ If p=3n–1:
p² = (3n – 1)² = 9n² – 6n + 1 = 3(3n² – 2n) + 1
and p² is logically 1 more than a multiple of 3.

So we have proved that any perfect square is:
either a multiple of 3,
or one more than a multiple of 3.

But what I don't understand is: 0 and 1 are also perfect squares but they are not multiples of 3. Then how can we say that every perfect square is a multiple of 3 or one more than a multiple of 3? maybe it's when we set p = 3n is whats confusing me.

Last edited: Mar 27, 2013
2. Mar 27, 2013

### HallsofIvy

Staff Emeritus
0 is a "multiple of three"- it is 3(0). And 1 is, of course, "one more than a multiple of 3".

3. Mar 27, 2013

### jfgobin

Hello John,

$0 = 3\cdot 0$, so $0$ is a multiple of $3$.

$1 = 0 + 1 = 3\cdot 0 + 1$

J.

4. Mar 27, 2013

### John112

So in a way does it mean:

3k + 1 = a number that is 1 bigger than a factor of 3 or a number that is two less than a factor of 3?

3k + 2 = a number that is 2 bigger than a factor of 3 or a number that is one less than a factor of 3?

5. Mar 27, 2013

### Curious3141

Yes. 1 bigger than a multiple of 3 is two less than the next higher multiple of 3.

Ditto for the second one.

Last edited: Mar 27, 2013
6. Mar 27, 2013

### skiller

Multiple!

7. Mar 27, 2013

### John112

So in a way does it mean:

3k + 1 = a number that is 1 bigger than a factor of 3 or a number that is two less than a factor of 3?

3k + 2 = a number that is 2 bigger than a factor of 3 or a number that is one less than a factor of 3?

For k $\geq$ 0,
so using n = 3k , we create all numbers like 0,3,6,9,12,15,18,...
using n = 3k + 1: we create all numbers like 1,4,7,10,13,16,19...
using n = 3k + 2: we create all numbers like 2,5,8,11,14,17,20,...

For k < 0,
using n = -3k + 2, we create all numbers like -1, -4, -7, ...
using n = -3k + 1, we create all numbers like -2, -5, -8, ...
using n = -3k, we create all numbers like -3, -6, -9, ...

Therefore we have all the possible numbers such that n $\in$ $Z$.

So,$n^{2}$ $\Rightarrow$ We are finding the perfect square of all integers.

Therefore, every perfect square is a multiple of 3 or one more than multiple of 3 because we created such a function f(n) = 3k + C such that all perfect squares were created as multiples of
3 plus a constant.

Am I correct in my reasoning?

8. Mar 27, 2013

### skiller

Multiple!

9. Mar 27, 2013

### Curious3141

Heh, brain fart.

10. Mar 27, 2013

### Curious3141

OK so far, except we both seem to be afflicted by a strange malady where we refer to multiples as factors. I'm blaming you - I seem to have caught it from you.

Not quite correct. For one thing, if k is already negative, you don't have to change the expression to "-3*k", etc. since that will just make the whole thing positive again. For another (this is not wrong, just redundant), you really you don't have to bother with the negative integers here - since the positive square roots of the forms 3k, 3k+1 and 3k-1 (or 3k+2, which is equivalent) would already have covered all the cases. What extra info are you adding by considering the negative square roots? When they're squared, they'll equal the same thing.

Or we're finding all the forms a perfect square can take, since a perfect square is, by definition, the square of an integer.

This part is unclear, and unnecessary. Why talk about introducing functions and "plus a constant", etc.? The foregoing is already clear enough.