I'm looking to answer this question Prove $\forall k\in\mathbb{Z}$, $3|k-2$ implies $3|k^2-1$. I'm not sure what to do. I'm trying to study but now I am getting stuck on these questions that don't give a lot of information. Thanks for the help!
Asked
Active
Viewed 81 times
1
-
22k-1=2(k-2)+4-1=2(k-2)+3 – Kaladin Aug 26 '14 at 05:53
-
This is quite trivial. since $k=3x+2$, we have $(3x+2)^2-1=9x^2+12x+3$, which is always divisible by $3$. – Golbez Aug 26 '14 at 05:58
-
1If $3\mid k-2$, then also $3\mid k+1$, because $k+1=(k-2)+3$. – Jyrki Lahtonen Aug 26 '14 at 06:05
3 Answers
3
A start: If $3$ divides $k-2$, then $k-2=3q$ for some integer $q$, and therefore $k=3q+2$.
Now compute $k^2-1$, that is, $(3q+2)^2-1$.
André Nicolas
- 507,029
3
$$3|(k-2)\implies3|(k-2+3)\implies3|(k+1)\implies3|(k+1)(k-1)\implies3|(k^2-1)$$
barak manos
- 43,109
-
Wow, I got more up-votes than the Jon Skeet of math stack exchange. – barak manos Aug 26 '14 at 06:26
-
0
If there is a counterexample to the assertion, i.e. if there exist an integer $k$ and integers $q, q', r \neq 0$ such that $$k-2 = 3q,\ k^{2} - 1 = 3q' + r,\ r = 1, 2,$$ then we have $$(3q+2)^{2} - 1 = 3q' + r,$$ so that $$9q^2 + 12q + 3(1-q') = r.$$ But $r = 1, 2$ is not divisible by $3$, qed.
Yes
- 20,719