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- #1

- Feb 14, 2012

- 3,595

- Thread starter anemone
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- Thread starter
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- #1

- Feb 14, 2012

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- Aug 30, 2012

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Since you got the problem stuck into my head, this is my revenge!

IT'S A SMALL WORLD AFTER ALL IT'S A SMALL WORLD AFTER ALL IT'S A SMALL WORLD AFTER ALL IT'S A SMALL SMALL WORLD!!!

Now that that's stuck in

-Dan

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- #3

- Feb 7, 2012

- 2,679

Since you got the problem stuck into my head, this is my revenge!

IT'S A SMALL WORLD AFTER ALL IT'S A SMALL WORLD AFTER ALL IT'S A SMALL WORLD AFTER ALL IT'S A SMALL SMALL WORLD!!!

Now that that's stuck inyourhead I can finally relax.

-Dan

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- #4

- Feb 7, 2012

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Thus there are altogether six possible solutions, as follows: $$ \begin{array}{c|c|c}(a,b)&\frac{a^2+b}{b^2-a} & \frac{b^2+a}{a^2-b} \\ \hline (1,2) & 1 & -5 \\ (2,1) & -5&1 \\ (2,2) & 3&3 \\ (2,3) & 1& 11 \\ (3,2) & 11 & 1 \\ (3,3) & 2& 2 \end{array}$$

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- Feb 14, 2012

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Case 1: $a=b$. Substituting, we have

$\dfrac{a^2+a}{a^2-a}=\dfrac{a+1}{a-1}=1+\dfrac{2}{a-1}$, which is an integer iff $(a-1)|2$. As $a>0$, the only possible values are $a-1=1 \text{or} 2$. Hence, $(a,\,b)=(2,\,2)$ or $(3,\,3)$.

Case 2: $a=b-1$. Substituting, we have

$\dfrac{b^2+a}{a^2-b}=\dfrac{(a+1)^2+a}{a^2-(a+1)}=\dfrac{a^2+3a+1}{a^2-a-1}=1+\dfrac{4a+2}{a^2-a-1}$

Once again, notice that $4a+2>0$ and hence for $\dfrac{4a+2}{a^2-a-1}$ to be an integer, we must have $4a+2\ge a^2-a-1$, that is $a^2-5a-3\le 0$. Hence, since $a$ is an integer, we can bound $a$ by $1\le a \le 5$. Checking all ordered pairs, we find that only $(a,\,b)=(1,\,2)$ or $(2,\,3)$ satisfy the given conditions.

Thus, the ordered pairs that work are $(a,\,b)=(2,\,2),\,(3,\,3),\,(1,\,2),\,(2,\,3),\,(2,\,1),\,(3,\,2)$