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 April 11th, 2018, 12:38 AM #1 Member   Joined: Apr 2018 From: On Earth Posts: 34 Thanks: 0 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? Thanks. April 11th, 2018, 01:04 AM   #2
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 Originally Posted by Student2018 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.

Last edited by Maschke; April 11th, 2018 at 01:15 AM. April 12th, 2018, 12:22 AM   #3
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 Originally Posted by Maschke 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? April 12th, 2018, 02:49 AM   #4
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 Originally Posted by Student2018 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. Tags multiple, numbers, square Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post shunya Algebra 12 March 30th, 2014 06:44 AM bash Number Theory 2 March 26th, 2013 02:46 PM aroratushar Algebra 6 August 18th, 2010 03:57 AM thatguy512 Algebra 1 December 1st, 2009 11:48 PM brunojo Number Theory 2 November 19th, 2007 08:55 AM

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