1. ## Divisibility (polynomials)

Hi! I've got a problem with this assignment. I really hope you can help me.

Find all such pairs of different natural numbers $(a,b)$ that

$b^2 + ab + 4$

is divisible by

$a^2 + ab + 4$

Thanks!

2. Originally Posted by PaulinaAnna
Hi! I've got a problem with this assignment. I really hope you can help me.

Find all such pairs of different natural numbers $(a,b)$ that

$b^2 + ab + 4$

is divisible by

$a^2 + ab + 4$
This looks like an interesting problem. If $a^2+ab+4$ divides $b^2+ab+4$ then it also divides $(b^2+ab+4) - (a^2+ab+4) = b^2-a^2,$ and hence it divides $a(b^2-a^2) = ab^2 - a^3$. But $ab^2 - a^3 = (a^2+ab+4)(b-a) -4(b-a)$, and therefore $a^2+ab+4$ must divide $4(b-a)$.

But then $ab < a^2+ab+4 \leqslant 4(b-a) <4b$. Thus $a<4$, and we only have to investigate the possibilities a = 1, 2 or 3.

When a = 1, the condition that $a^2+ab+4$ divides $4(b-a)$ becomes: $b+5$ divides $4(b-1) = 4(b+5) - 24$, so that $b+5$ must be a factor of 24. You can then check that there are only three such values for b (given that b is different from a). Similar calculations (I'll leave you to work out the details) show that there are no solutions for a=2, and only one solution for a=3.

Spoiler:
The solutions are $(a,b) = (1,3),\ (1,7),\ (1,19)$ and $(3,25)$.

3. I've got a problem with a=3. I get $a^2+ab+4=3b+13$ and $b-a=b-3$, so $3b+13|4(b-3)$. What do I do now? And how do I prove that there are no solutions for a=2?

4. Originally Posted by PaulinaAnna
I've got a problem with a=3. I get $a^2+ab+4=3b+13$ and $b-a=b-3$, so $3b+13|4(b-3)$. What do I do now? And how do I prove that there are no solutions for a=2?
If 3b+13 divides 4(b–3) then it divides 12(b–3) = 4(3b+13) – 88, so it divides 88. The only divisors of 88 which are larger than 13 are 22, 44 and 88. The only one of those that gives a solution to the problem is 88 = 3x25 + 13, so b=25.

For a=2, you should get the condition that 2b+8 divides 4(b–2). Therefore b+4 divides 2(b–2) = 2(b+4) – 12. So b+4 divides 12. The only divisors of 12 which are greater than 4 are 6 and 12, and neither of these gives a solution to the problem.

5. ## Thank you

Thank you again very much.

6. Wow.. congrats for solving Opalg, it was quite a bit of a problem! However, I don't get one thing

Originally Posted by Opalg
But then $ab < a^2+ab+4 \leqslant 4(b-a) <4b$. Thus $a<4$, and we only have to investigate the possibilities a = 1, 2 or 3.
Where do know all this inequality from? How do we know which part is lesser than the other?

7. Originally Posted by Glyper
Wow.. congrats for solving Opalg, it was quite a bit of a problem! However, I don't get one thing

Originally Posted by Opalg
But then $ab < a^2+ab+4 \leqslant 4(b-a) <4b$. Thus $a<4$, and we only have to investigate the possibilities a = 1, 2 or 3.

Where do know all this inequality from? How do we know which part is lesser than the other?
In that string of inequalities, the first one $ab < a^2+ab+4$ is obvious, the second one $a^2+ab+4 \leqslant 4(b-a)$ comes from the fact that $a^2+ab+4$ divides $4(b-a)$ (and hence must be smaller than it), and the final one $4(b-a) <4b$ is again obvious. When you string all the inequalities together, you see that $ab<4b$ and therefore (dividing by $b$) $a<4$.

8. Originally Posted by Opalg
For a=2, you should get the condition that 2b+8 divides 4(b–2). Therefore b+4 divides 2(b–2) = 2(b+4) – 12. So b+4 divides 12. The only divisors of 12 which are greater than 4 are 6 and 12, and neither of these gives a solution to the problem.
Why? If b+4=12, b=8 and 12 divides 84 so it should be a solution, shouldn't it?

9. Originally Posted by Glyper
Originally Posted by Opalg
For a=2, you should get the condition that 2b+8 divides 4(b–2). Therefore b+4 divides 2(b–2) = 2(b+4) – 12. So b+4 divides 12. The only divisors of 12 which are greater than 4 are 6 and 12, and neither of these gives a solution to the problem.
Why? If b+4=12, b=8 and 12 divides 84 so it should be a solution, shouldn't it?
We are looking for pairs of numbers $a,\ b$ such that $a^2+ab+4$ divides $b^2+ab+4$. If $a=2$ and $b=8$ then $a^2+ab+4 = 24$ and $b^2+ab+4 = 84$. Unfortunately 24 does not divide 84. So although these numbers look like a promising pair, they do not satisfy the conditions of the original problem.

10. Thank you a lot