$k$ divides the product of $k$ consecutive integers (case $k=2,3)$

Question

How do I show that for any arbitrary integer $a$, $2|(a+1)$ and $3a|(a+1)(a+2)$?

$2|(a+1)$, if $a$ is even, $2k=a$ replacing $a$ , $2k=a$ for some $k$ in $Z$

\begin{align}a(a+1)&=2k(2k+1)\\ &=4k^2+2k\\ &=2(2k^2+k)\end{align}

shows clearly that $2$ is divisible $(a+1)$

but I can't prove the second part

Could you please try to express more clearly your actual problem. — Angina Seng, May 13 '17 at 18:28
It is not true that $2\mid (a+1)$ for "any arbitrary integer $a$". You only showed that $2\mid a(a+1)$. The second part is also false - take $a=5$. — Dietrich Burde, May 13 '17 at 18:40
Did you want to say $3\mid a(a+1)(a+2)$? (One of those factors is a multiple of $3$.) — Angina Seng, May 13 '17 at 18:45
@LordSharktheUnknown Maybe he meant $2|a(a+1)$ and $3|a(a+1)(a+2)$. — u8y7541, May 13 '17 at 19:02
@u8y7541 I suspect he/she does. But I was hoping he/she would edit the question to make this clear. — Angina Seng, May 13 '17 at 19:15
@LordSharktheUnknown Well, the first part must have been $2|a(a+1)$ because that's how he solves it. — u8y7541, May 13 '17 at 19:55
$2 \not \mid a+1$ for any arbitrary integer as if $a$ is even then $a+1$ is odd and $2 \not \mid a+1$. I suspect you meant $2| a(a+1)$ which is true. But you only provd it for and even $a$. You must also prove it for every $a$ even the $a$ that aren't even. And $3a \not \mid (a+1)(a+2)$ unless $a = 1$. — fleablood, May 13 '17 at 20:48
"because that's how he solves it. " But he only solves it for even a. — fleablood, May 13 '17 at 20:48
There is a typo, it should be $, 2\mid a(a+1)., $ See the linked dupe for the general case. — Bill Dubuque, May 13 '17 at 21:56
There are typos. it should be $2∣a(a+1)$ and $,3\mid a(a+1)(a+2).,$ See the linked dupe for the general case — Bill Dubuque, May 13 '17 at 22:19

score 1 · Answer 1 · answered May 13 '17 at 20:44

1

Both statements are actually false.

for $a|(a+1)$ take $a = 2$. Clearly $2$ does not divide $3$

for $3a|(a+1)(a+2)$, take $a = 5$. Clearly $15$ does not divide $6*7 = 42$

answered May 13 '17 at 20:44

Actually $3a|(a+1)(a+2)=a^2 + 3a + 2\iff a|2$ and $3|a^2 + 2$. The only possible solution are a = 1 or a=2 – fleablood May 13 '17 at 20:54
The statement in the OP says for arbitrary $a$ – May 13 '17 at 20:56
Um... yeah? So it is definitely false for an arbitrary $a$ because it is only true for two $a$... – fleablood May 13 '17 at 21:12

$k$ divides the product of $k$ consecutive integers (case $k=2,3)$

1 Answers1