Hi! I've got a problem with this assignment. I really hope you can help me.
Find all such pairs of different natural numbers that
is divisible by
Thanks!
This looks like an interesting problem. If $\displaystyle a^2+ab+4$ divides $\displaystyle b^2+ab+4$ then it also divides $\displaystyle (b^2+ab+4) - (a^2+ab+4) = b^2-a^2,$ and hence it divides $\displaystyle a(b^2-a^2) = ab^2 - a^3$. But $\displaystyle ab^2 - a^3 = (a^2+ab+4)(b-a) -4(b-a)$, and therefore $\displaystyle a^2+ab+4$ must divide $\displaystyle 4(b-a)$.
But then $\displaystyle ab < a^2+ab+4 \leqslant 4(b-a) <4b$. Thus $\displaystyle a<4$, and we only have to investigate the possibilities a = 1, 2 or 3.
When a = 1, the condition that $\displaystyle a^2+ab+4$ divides $\displaystyle 4(b-a)$ becomes: $\displaystyle b+5$ divides $\displaystyle 4(b-1) = 4(b+5) - 24$, so that $\displaystyle b+5$ 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:
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.
In that string of inequalities, the first one $\displaystyle ab < a^2+ab+4$ is obvious, the second one $\displaystyle a^2+ab+4 \leqslant 4(b-a)$ comes from the fact that $\displaystyle a^2+ab+4$ divides $\displaystyle 4(b-a)$ (and hence must be smaller than it), and the final one $\displaystyle 4(b-a) <4b$ is again obvious. When you string all the inequalities together, you see that $\displaystyle ab<4b$ and therefore (dividing by $\displaystyle b$) $\displaystyle a<4$.
We are looking for pairs of numbers $\displaystyle a,\ b$ such that $\displaystyle a^2+ab+4$ divides $\displaystyle b^2+ab+4$. If $\displaystyle a=2$ and $\displaystyle b=8$ then $\displaystyle a^2+ab+4 = 24$ and $\displaystyle b^2+ab+4 = 84$. 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.