Question

Alright, here is the Fooool's problem of the day:

If for some natural numbers \( A\) and \( B\), \( B^2 +BA+1\) divides \(A^2 +AB +1\),
then \( A − B = k\), where \( k \) is an integer. How many such \( k\) exist?

Answer 1

Am i? To me I look as red as before.

Answer 2

Duh, I just realized that this is the easiest one of all I posted.

Answer 3

Sorry not the correct answer.

PS: Does programming fails after all? :P

Answer 4

I am nowhere near the solution.

Answer 5

And, I am sure this isn't the easiest of all foolformath's questions.

or maybe, I am missing something.

Answer 6

If it helps, I am getting a crazy expression after some substitution and algebra, which can be incorrect too.

\[1 + \frac{2kB + k^2}{2B^2 + Bk +1}=n, n \in \mathbb{R^{+}}\]

Answer 7

Ishaan forced me to sign in to say don't give hints just yet. I'm getting the answer :P

Answer 8

-sigh- I wish I was a mathematical Genius, do all kinds of calculation and algebra in my head.

Sorry lol :-D

Answer 9

Ishaan, this is not typical algebra.

Answer 10

I think A has to be equal to B, hence there's only one such k and k=0.

Answer 11

But what about others? :-D


btw I think your solution is right.

Answer 12

I do have a proof by the way (or so I think). :D

Answer 13

Yes, that's right. Let us see the proof Mr.Math.

Answer 14

Sure! :)

Answer 15

Alright! First, someone can easily see that \(A^2+AB+1>0\) and \(B^2+AB+1>0\), for \(A, B\in \mathbb{N}\).
So it follows that \(A^2+AB+1\ge B^2+AB+1 \implies A\ge B\), and we can write:
\[cA^2+cAB+c=B^2+AB+1, \text{ where c is an integer}\ge 1. \]
From that last equation we can get:
\[(c-1)A^2+A^2-B^2+(c-1)AB+c-1=0\]
\[\implies (c-1)(A^2+AB+1)+(A-B)(A+B)=0,\]
\[ \text{ but since A-B=k, we can write}:\]
\[k=\frac{(1-c)(A^2+AB+1)}{A+B}.\]
That's (A+B)|(A^2+AB+1) or (A+B)|(1-c):
Since \(\frac{A^2+AB+1}{A+B}=\frac{1}{A+B}+1\), then there's no natural numbers A,B such that (A+B)|(A^2+AB+1). So, it has to be the case that (A+B)|(1-c). The obvious solution here is that 1-c=0 or c=1.
Hmm, I think there's still something missing here.

Answer 16

Yeah, I got what's missing. We have shown above that A>=B, that k>=0, which implies that \[1-c\ge 0 \implies c\le 1\]
Thus c=1 is the only solution, which gives k=0.

Answer 17

Is the proof clear?

Answer 18

You lost me after the first line itself :P

Answer 19

A and B are natural then A^2+AB>=1, right?

Answer 20

In other words both the top and bottom are positive, so we need not to worry about negative divisors (i.e the top has to be greater than or equal to the bottom).

Answer 21

And it follows that A>=B.

Answer 22

Yes yes, I mean't this was not the elementary proof that I was talking about :)

Answer 23

Oh, I see. I know I always complicate things :P

Answer 24

It's so simple, If A divides B then A divides B-A, and then ... ;)

Answer 25

I like my proof! :P

Answer 26

Ishaan likes it too (I think).

Answer 27

If I post mine yours will be rendered too complicated :P

Answer 28

Maybe! :D

Answer 29

Ahh, I have to get back to study group theory! :(

Answer 30

Honestly, I didn't know (or consider) that if A|B then A|(B-A).

Answer 31

That's Number theory 101 for any Mathelete ;)

Answer 32

Yeah, when I think about it. Its proof is trivial too.