Subfield: Number Theory


Question: Suppose that $c$ is a positive integer. Define $f(c)$ to be the number of pairs $(a, b)$ of positive integers with $c<a<b$ for which two circles of radius $a$, two circles of radius $b$, and one circle of radius $c$ can be drawn so that

- each circle of radius $a$ is tangent to both circles of radius $b$ and to the circle of radius $c$, and

- each circle of radius $b$ is tangent to both circles of radius $a$ and to the circle of radius $c$,

as shown. Determine all positive integers $c$ for which $f(c)$ is even.


![](https://cdn.mathpix.com/cropped/2023_12_21_dff3b889a8199883de5cg-1.jpg?height=395&width=436&top_left_y=1426&top_left_x=1235)

Solution:
Let the centre of the middle circle be $O$, and the centres of the other circles be $P, Q, R$, and $S$, as shown.

Join $O$ to $P, Q, R$, and $S$, and join $P$ to $Q, Q$ to $R, R$ to $S$, and $S$ to $P$.



![](https://cdn.mathpix.com/cropped/2023_12_21_15aa73ce70e6e4807dc7g-1.jpg?height=675&width=745&top_left_y=156&top_left_x=793)

As the label in the diagram, we see that

$$
\begin{array}{r}
O P=O R=a+c \\
O Q=O S=b+c \\
P Q=Q R=R S=S P=a+b
\end{array}
$$

By side-side-side congruence, $\triangle O P Q, \triangle O P S, \triangle O R Q$, and $\triangle O R S$ are congruent.

This means that $\angle P O Q=\angle P O S=\angle R O Q=\angle R O S$.

Since $\angle P O Q+\angle P O S+\angle R O Q+\angle R O S=360^{\circ}$ (these angles surround $O$ ), then

$$
\angle P O Q=\frac{1}{4} \cdot 360^{\circ}=90^{\circ}
$$

This means that $\triangle O P Q$ is right-angled at $O$.

By the Pythagorean Theorem, $P Q^{2}=O P^{2}+O Q^{2}$ and so $(a+b)^{2}=(a+c)^{2}+(b+c)^{2}$. Manipulating algebraically, the following equations are equivalent:

$$
\begin{aligned}
(a+b)^{2} & =(a+c)^{2}+(b+c)^{2} \\
a^{2}+2 a b+b^{2} & =a^{2}+2 a c+c^{2}+b^{2}+2 b c+c^{2} \\
2 a b & =2 a c+2 b c+2 c^{2} \\
a b & =a c+b c+c^{2} \\
a b-a c-b c & =c^{2} \\
a b-a c-b c+c^{2} & =2 c^{2} \\
a(b-c)-c(b-c) & =2 c^{2} \\
(a-c)(b-c) & =2 c^{2}
\end{aligned}
$$

Therefore, if $a, b$ and $c$ are real numbers for which the diagram can be constructed, then $a, b$ and $c$ satisfy this last equation.

Also, if real numbers $a, b$ and $c$ satisfy the final equation, then $(a+b)^{2}=(a+c)^{2}+(b+c)^{2}$ (because these equations were equivalent) and so the triangle with side lengths $a+b$, $a+c$ and $b+c$ is right-angled with hypotenuse $a+b$ (because the Pythagorean Theorem works in both directions), which means that four such triangles can be assembled to form a rhombus $P Q R S$ with side lengths $a+b$ and centre $O$, which means that the five circles can be drawn by marking off the appropriate lengths $a, b$ and $c$ and drawing the circles as in the original diagram.



In other words, the diagram can be drawn exactly when $(a-c)(b-c)=2 c^{2}$.

Suppose that $c$ is a fixed positive integer.

Determining the value of $f(c)$ is thus equivalent to counting the number of pairs of positive integers $(a, b)$ with $c<a<b$ and $(a-c)(b-c)=2 c^{2}$.

Since $a$ and $b$ are integers with $a>c$ and $b>c$, the integers $a-c$ and $b-c$ are positive and form a positive divisor pair of the integer $2 c^{2}$.

Since $a<b$, we have $a-c<b-c$ and so $a-c$ and $b-c$ are distinct integers.

Also, since $c>0, \sqrt{2 c^{2}}=\sqrt{2} c$ which is not an integer since $c$ is an integer, which means that $2 c^{2}$ is not a perfect square.

Therefore, every pair $(a, b)$ corresponds to a positive divisor pair of $2 c^{2}$ (namely, $a-c$ and $b-c)$.

Similarly, every divisor pair $e$ and $g$ of $2 c^{2}$ with $e>g$ gives a pair of positive integers $(a, b)$ with $a<b$ by setting $a=e+c$ and $b=g+c$.

In other words, $f(c)$ is exactly the number of positive divisor pairs of $2 c^{2}$. (Again, we note that $2 c^{2}$ is not a perfect square.)

Therefore, we want to determine all positive integers $c$ for which the integer $2 c^{2}$ has an even number of divisor pairs, which means that we want to determine all positive integers $c$ for which the number of positive divisors of $2 c^{2}$ is a multiple of 4 (because each positive divisor pair corresponds to 2 positive divisors and 2 times an even integer is a multiple of 4$)$.

Suppose that the prime factorization of $c$ is

$$
c=2^{r} p_{1}^{e_{1}} p_{2}^{e_{2}} \cdots p_{k}^{e_{k}}
$$

for some integer $k \geq 0$, integer $r \geq 0$, odd prime numbers $p_{1}, p_{2}, \ldots, p_{k}$, and positive integers $e_{1}, e_{2}, \ldots, e_{k}$.

Then

$$
2 c^{2}=2^{2 r+1} p_{1}^{2 e_{1}} p_{2}^{2 e_{2}} \cdots p_{k}^{2 e_{k}}
$$

and so $2 c^{2}$ has

$$
(2 r+2)\left(2 e_{1}+1\right)\left(2 e_{2}+1\right) \cdots\left(2 e_{k}+1\right)
$$

positive divisors.

The first factor in this product is even and each factor after the first is odd.

Therefore, this product is a multiple of 4 exactly when $2 r+2$ is a multiple of 4 .

This is true exactly when $2 r+2=4 s$ for some positive integer $s$ and so $2 r=4 s-2$ or $r=2 s-1$.

In other words, the number of positive divisors of $2 c^{2}$ is a multiple of 4 exactly when $r$ is an odd integer.

Finally, this means that the positive integers for which $f(c)$ are even are exactly those positive integers that have exactly an odd number of factors of 2 in their prime factorization.

Final answer: the positive integers for which $f(c)$ are even are exactly those positive integers that have exactly an odd number of factors of 2 in their prime factorization