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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  
Senior Member Joined: Aug 2012 Posts: 1,847 Thanks: 507  Quote:
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  
Member Joined: Apr 2018 From: On Earth Posts: 34 Thanks: 0  Quote:
 
April 12th, 2018, 02:49 AM  #4  
Senior Member Joined: Oct 2009 Posts: 350 Thanks: 113  Quote:
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 $n=3m1$. Then, $n^2 = (3m1)^2 = 9m^2  6m +1 = 3(3m^22m)+1$ 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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