If a is odd, prove ... by division algorithm

**janchilds** · September 9th 2012, 08:52 AM

The problem states that if a is odd, then prove 12 | a^2 + (a+2)^2 + (a+4)^2 + 1

Here is what I did but it seems more by induction than the division algorithm. If someone could let me know if I am on the right track.

Base case a=1 .... 12 |36 done.

So, if k is in G then prove (2k-1) is in G

12 | (2k-1)^2 + ((2k-1)+2)^2 + ((2k-1)+4)^2 +1 - I simply multiplied this whole thing out and got 12 | 12k^2 + 12k + 12

Then if a|b we know a|bc, so 12|12 or 12|-12 if we want to use if a|b and a|c then a|b-c

Thus k+1 is in G.

I guess I just thought I would be looking for k(k+1)(k+2) as we did in the example in class.

**MaxJasper** · September 9th 2012, 09:16 AM

a=odd = $a\text{=}2k+1$

$a^2+(a+2)^2+(a+4)^2+1 = 12 \left(3+3 k+k^2\right)$

**Soroban** · September 9th 2012, 09:33 AM

Hello, janchilds!

You are trying an inductive proof . . . Okay!

$\text{If }a\text{ is odd, then }a^2 + (a+2)^2 + (a+4)^2 + 1\text{ is divisible by 12.}$

Verify $S(1)\!:\;1^2 + 3^2 + 5^2 + 1 \:=\:36\;\hdots\;\text{True!}$

Assume $S(k)\!:\;k^2 + (k+2)^2 +(k+4)^2 + 1 \;=\;12p\;\text{ for some integer }p.$

Add $(k+6)^2-k^2$ to both sides:

. . $(k+2)^2 + (k+4)^2 + (k+6)^2 + 1 \;=\;12p + (k+6)^2-k^2$

. . $(k+2)^2 + (k+4)^2 + (k+6)^2 + 1 \;=\;12p + k^2 +12k + 36 - k^2$

. . $(k+2)^2 + (k+4)^2 + (k+6)^2 + 1 \;=\;12p + k + 36$

. . $(k+2)^2 + (k+4)^2 + (k+6)^2 + 1 \;=\;\underbrace{12(p + k + 3)}_{\text{divisible by 12}}$

We have proved $S(k\!+\!1).$
The inductive proof is complete.

**janchilds** · September 9th 2012, 11:55 AM

I get this... a bit different format from what I was doing but the same bottom line.

But this is proving it by Induction. If the instructions specifically says prove using the Division Algorithm.... is that the same?

Thanks,
Janet

**SlipEternal** · September 9th 2012, 12:14 PM

You can add the division algorithm into the proof rather easily. Write the statement of the proof, then any time you are discussing divisibility, you just need $r=0$ where $r$ is the remainder in the division algorithm. So, your proof by induction changes form. You claim that for any $a\in 2\mathbb{Z}+1$ , there exists $p\in \mathbb{Z}$ so that $(a^2+(a+2)^2+(a+4)^2+1) = 12p + 0$ . Now, you are using the division algorithm where your $q=12$ and $r=0$ .

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