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Thread: Divisibility (polynomials)

  1. #1
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    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 that



    is divisible by



    Thanks!
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    Quote Originally Posted by PaulinaAnna View Post
    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

    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:
    The solutions are $\displaystyle (a,b) = (1,3),\ (1,7),\ (1,19)$ and $\displaystyle (3,25)$.
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    I've got a problem with a=3. I get $\displaystyle a^2+ab+4=3b+13$ and $\displaystyle b-a=b-3$, so $\displaystyle 3b+13|4(b-3)$. What do I do now? And how do I prove that there are no solutions for a=2?
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    Quote Originally Posted by PaulinaAnna View Post
    I've got a problem with a=3. I get $\displaystyle a^2+ab+4=3b+13$ and $\displaystyle b-a=b-3$, so $\displaystyle 3b+13|4(b-3)$. What do I do now? And how do I prove that there are no solutions for a=2?
    If 3b+13 divides 4(b3) then it divides 12(b3) = 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(b2). Therefore b+4 divides 2(b2) = 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.
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    Thank you

    Thank you again very much.
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    Wow.. congrats for solving Opalg, it was quite a bit of a problem! However, I don't get one thing

    Quote Originally Posted by Opalg View Post
    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.
    Where do know all this inequality from? How do we know which part is lesser than the other?
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    Quote Originally Posted by Glyper View Post
    Wow.. congrats for solving Opalg, it was quite a bit of a problem! However, I don't get one thing

    Quote Originally Posted by Opalg View Post
    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.

    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 $\displaystyle ab < a^2+ab+4$ 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 $\displaystyle b$) $\displaystyle a<4$.
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    Quote Originally Posted by Opalg View Post
    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?
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    Quote Originally Posted by Glyper View Post
    Quote Originally Posted by Opalg View Post
    For a=2, you should get the condition that 2b+8 divides 4(b2). Therefore b+4 divides 2(b2) = 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 $\displaystyle a,\ b$ 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.
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    Thank you a lot
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