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April 11th, 2018, 01:38 AM   #1
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Why is it that none of the square numbers is one less than a multiple of 3?

I've noticed that none of the square numbers is one less than a multiple of 3. I have a rough idea of why this is but I'm not completely sure. Please can someone clarify?

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April 11th, 2018, 02:04 AM   #2
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I've noticed that none of the square numbers is one less than a multiple of 3. I have a rough idea of why this is but I'm not completely sure. Please can someone clarify?
Yes, it's modular arithmetic. A number that's one less than a multiple of 3 is -1 mod 3. Or 2 mod 3, comes out to the same thing. In other words if some number $n$ can be written as $3 m - 1$, by adding another multiple of $3$ we can write the same number as $3 (m - 1) + 2$.

Symbolically we write $n \equiv -1 \pmod 3$ or $n \equiv 2 \pmod 3$. Either way depending on what you need at the moment.

Now what are the square numbers mod 3?

* $0^2 \equiv 0 \pmod 3$.

* $1^2 \equiv 1 \pmod 3$.

* $2^2 \equiv 1 \pmod 3$.

There are no other possibilities and none of these are 2. So one less than a multiple of 3 can never be a perfect square.

Once you get the hang of this you can solve all kinds of similar problems.
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Last edited by Maschke; April 11th, 2018 at 02:15 AM.
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April 12th, 2018, 01:22 AM   #3
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Originally Posted by Maschke View Post
Yes, it's modular arithmetic. A number that's one less than a multiple of 3 is -1 mod 3. Or 2 mod 3, comes out to the same thing. In other words if some number $n$ can be written as $3 m - 1$, by adding another multiple of $3$ we can write the same number as $3 (m - 1) + 2$.

Symbolically we write $n \equiv -1 \pmod 3$ or $n \equiv 2 \pmod 3$. Either way depending on what you need at the moment.

Now what are the square numbers mod 3?

* $0^2 \equiv 0 \pmod 3$.

* $1^2 \equiv 1 \pmod 3$.

* $2^2 \equiv 1 \pmod 3$.

There are no other possibilities and none of these are 2. So one less than a multiple of 3 can never be a perfect square.

Once you get the hang of this you can solve all kinds of similar problems.
Is there any way to solve this without using mods, but just by explaining one's reasoning?
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April 12th, 2018, 03:49 AM   #4
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Is there any way to solve this without using mods, but just by explaining one's reasoning?
Yes of course, since mods are just a shorthand. I do encourage you to learn mods, since it is a very useful tool and not hard to learn.

In any case. For a number $n$, there are three possibilities:
1) $n$ is divisible by $3$, hence $n=3m$.
Then $n^2=9m^2$ is also divisible by three, so can't be one less than a multiple of three.

2) $n$ is one less than a multiple of three, hence $n=3m-1$.
Then, $n^2 = (3m-1)^2 = 9m^2 - 6m +1 = 3(3m^2-2m)+1$
This is one more than a multiple of three, so can't be one less than a multipple of three.

3) $n$ is one more than a multiple of three, hence $n=3m+1$.
Then $n^2=(3m+1)^2= 9m^2+6m + 1 = 3(3m^2+2m)+1$.
This is one more than a multiple of three, so again: it can't be one less than a multiple of three.

So given any number, its square is never less than a multiple of three. Done.
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