$n^2 +1$ is not divisible by 3 for any integer n

Asked Mar 24 '21 at 20:59

Active Mar 25 '21 at 01:15

Viewed 74 times

So I understand the basic concept of when $n=3k $
$(3k)^2 + 1 = 9k^2 + 1 = 3(3k^2) + 1 $

But I saw a solution when we use $3k+ 1$.
for example for $(3k+1)^2 + 1 = 9k^2 + 6k + 1 + 1 = 3(3k^2 + 2k) + 2.$
where does the 6 come from?

Same goes for $3k +2.$
and $(3k+2)^2 + 1 = 9k^2 + 12k + 4 + 1 = 3(3k^2 + 4k) + 3 + 2 = 3(3k^2 + 4k + 1) + 2. $ where does the 12 come from? How does the +4 and +1 change to +3 and +2? And then how do we get the + 1 back in the brackets at the end?

Thanks to anyone that can help!

edited Mar 25 '21 at 01:15

asked Mar 24 '21 at 20:59

Dani

1

Calculate $(3k+1)^2$ yourself and see what you get. – Arthur Mar 24 '21 at 21:01
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The $6$ and $12$ come from the standard expansion $(a + b)^2 = a^2 + 2ab + b^2$, which in turn comes from expanding $(a + b)(a + b)$. In this case, substitute $a = 3k$ and $b = 1$ (or $b = 2$) to see how the $6$ and $12$ come about. – Theo Bendit Mar 24 '21 at 21:01
ahhhh makes sense! Thank you! – Dani Mar 24 '21 at 21:02
$12k=2.3k.2. (a+b)^2=2ab+...$ – hamam_Abdallah Mar 24 '21 at 21:03
It is so much simpler to prove that using congruences and lil' Fermat… – Bernard Mar 24 '21 at 21:05
So then my other question would be how does the +4 and +1 change to + 3 and +2 in the following equation? and how did we get the +1 into the brackets? – Dani Mar 24 '21 at 21:13
Please edit your question using MathJax. – user Mar 24 '21 at 21:16
It is also true that $m^2 + n^2$ is not divisible by $3$ unless both $m,n$ are divisible by $3$ – Will Jagy Mar 24 '21 at 21:19
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What are the squares mod 3? That is, what are the values $0^2,1^2,$ and $2^2$ mod $3$? Is any one of them equal to $2$ mod $3?$ – Chris Leary Mar 24 '21 at 21:27
AHH ok makes sense now! – Dani Mar 24 '21 at 21:30

$n^2 +1$ is not divisible by 3 for any integer n

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