# Solving Integer Equations with $a$ and $b$

• MHB
• anemone
In summary, the conversation discusses a problem that has gotten stuck in one person's head and how they seek revenge by getting a song stuck in the other person's head. The problem involves finding positive integers $a$ and $b$ that satisfy a certain condition.
anemone
Gold Member
MHB
POTW Director
Find all positive integers $a$ and $b$ such that $\dfrac{a^2+b}{b^2-a}$ and $\dfrac{b^2+a}{a^2-b}$ are both integers.

This problem has gotten stuck in my head. This would be okay if I had any inkling of how to solve it. (I already went through all my ideas and they failed.) So this message is for anemone only. Open it at your own risk.

[sp]
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!

[/sp]

-Dan

topsquark said:
[sp]
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!

[/sp]
-Dan
I love that song! I first heard it in the Disney pavilion when I visited the New York World's Fair in 1965. Then in 1979 I took my kids to Disney World in Florida, and there it was again. Happy memories!

anemone said:
Find all positive integers $a$ and $b$ such that $\dfrac{a^2+b}{b^2-a}$ and $\dfrac{b^2+a}{a^2-b}$ are both integers.
The problem is symmetric in $a$ and $b$, so it will be sufficient to find all solutions with $a\leqslant b$. Let $b = a+c$, with $c\geqslant0$. Then $$\frac{a^2+b}{b^2-a} = \frac{a^2+a+c}{a^2 + (2c-1)a + c^2}, \qquad \frac{b^2+a}{a^2-b} = \frac{a^2 + (2c+1)a + c^2}{a^2-a-c} = 1 + \frac{(2c+2)a + c(c+1)}{a^2 - a - c}.$$ Case 1: $c=0$. Then both fractions become $\dfrac{a+1}{a-1}$, which is an integer when $a = 2$ or $3$ but not otherwise.

Case 2: $c=1$. Then the first fraction becomes $\dfrac{a^2+a+1}{a^2+a+1}$, which is obviously an integer for every $a$. The second fraction is $1 + \dfrac{4a+2}{a^2-a-1}$. That is an integer when $a = 1$ or $2$. It is not an integer if $a= 3,\ 4$ or $5$. And if $a\geqslant 6$ then $0 < \dfrac{4a+2}{a^2-a-1} < 1$. So there can be no more solutions in this case.

Case 3: $c>1$. Then in the fraction $\dfrac{a^2+a+c}{a^2 + (2c-1)a + c^2}$ the numerator is less than the denominator. So the fraction lies between $0$ and $1$ and is therefore not an integer.

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

By the symmetry of the problem we may suppose $a\le b$. Notice that $b^2-1\ge 0$ so that if $\dfrac{a^2+b}{b^2-a}$ is a positive integer, then $a^2+b\ge b^2-a$. Rearranging this inequality and factoring we find that $(a+b)(a-b+1)\ge 0$. Since $a,\,b>0$, we must have $a\ge b-1$. We therefore have two cases:

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 $(a,\,b)=(2,\,2),\,(3,\,3),\,(1,\,2),\,(2,\,3),\,(2,\,1),\,(3,\,2)$

## What are integer equations?

Integer equations are mathematical expressions that involve whole numbers, also known as integers. These equations typically contain variables, such as $a$ and $b$, and require solving for the unknown values.

## How do I solve integer equations with $a$ and $b$?

To solve integer equations with $a$ and $b$, you can use basic algebraic principles such as combining like terms, using inverse operations, and applying the distributive property. It is important to follow the order of operations and always check your solution by substituting the values back into the original equation.

## What is the difference between linear and quadratic integer equations?

Linear integer equations involve variables that are raised to the first power, while quadratic integer equations involve variables that are raised to the second power. This means that quadratic equations can have two solutions, while linear equations have only one solution.

## Can I use the same method to solve all integer equations?

No, the method used to solve integer equations may vary depending on the type of equation. For example, linear equations can be solved using the substitution or elimination method, while quadratic equations can be solved using factoring or the quadratic formula.

## Why is it important to check my solution when solving integer equations?

Checking your solution is important because it ensures that you have found the correct values for the variables. Sometimes, a solution may seem correct but may not satisfy the original equation. Checking also helps to catch any mistakes made during the solving process.

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