Sreejato's ordered pairs

Because all of $a,b,x$ are positive, we can multiply through by $a \cdot b \cdot x$ in the equation $\frac{1}{a} + \frac{1}{b} = \frac{1}{x}$ to get $x(a + b) = ab$ . Rearranging and factoring gives us $(a - x)(b - x) = x^2$ , and from this, it is clear that $f(x)$ is the number of divisors of $x^2$ , since $(a-x),(b-x) \in \mathbb{Z}$ .

We have $x = p^n$ with $p$ a positive prime and $n$ a positive integer. The previous equation becomes $(a - p^n)(b - p^n) = p^{2n}$ . Clearly, $p^{2n}$ has $2n + 1$ divisors, and so $f(p^n) = 13 \to 2n + 1 = 13 \to n = 6$ . Thus, we want the smallest possible value of $p^6$ where $p$ is a positive prime, which is obviously $2^6 = \fbox{64}$ .

I solved this problem with the assumption that $f(N)=13$ . I shall explain why later. The solutions to the Diophantine equation $\frac{1}{a} + \frac{1}{b} = \frac{1}{x}$ are related to the divisors of $x^2$ . This is because

$\frac{1}{a}+\frac{1}{b}=\frac{1}{x}$

$\frac{a+b}{ab}=\frac{1}{x}$

From this we have

$ab-ax-bx=0$

Adding $x^{2}$ to both sides

$x^{2}+ab-ax-bx=x^{2}$

Factorizing we get

$(a-x)(b-x)=x^{2}$

Hence we see that the number of solutions to $\frac{1}{a}+\frac{1}{b}=\frac{1}{x}$ is $\tau(x^{2})$ which means that $f(x)=\tau(x^{2})$ ,where $\tau(x)$ is a function that counts the number of divisors of $x$ . It is also well known that for an integer $N$ with prime factorization $N=p1^{a1} . p2^{a2} . p3^{a3} ...$ $\tau(N)=(a1+1).(a2+1).(a3+1)..$ .

Now we have:

$13=\tau(N^2)$

$13=(a1+1).(a2+2).(a3+3)...=\tau(N^2)=f(N)$

but since $N$ is a prime power its factorization should be of the form $N=p1^{a1+1}$

so:

$a1+1=13$

$a1=12$

$N^2=p^{12}$

$N=p^6$

If $N$ had been $12$ we would have got 6.5 as an exponent.

Since we are asked for the smallest possible value of $N$ we take the smallest possible prime $2$

Giving us $N=2^6=64$

We want to find the number of ordered pairs of positive integers $(a,b)$ such that

$\frac{1}{a} + \frac{1}{b} = \frac{1}{N}.$

Obviously, $a,b>N$ , because $a$ and $b$ are both positive. Let $b = N + d$ for some positive integer $d$ . The equation then becomes

$\frac{1}{a} + \frac{1}{N+d} = \frac{1}{N} \implies \frac{1}{a} = \frac{1}{N} - \frac{1}{N+d} = \frac{d}{N(N+d)}.$

So,

$a = \frac{N(N+d)}{d} = N + \frac{N^2}{d}.$

The number of ordered pairs $(a,b)$ that satisfy the equation is therefore the same as the number of values $d$ can take, i.e. the number of positive divisors of $N^2$ . We know that $N$ is a prime power, i.e.

$N = p^n \implies N^2 = p^{2n},$

and $p^{2n}$ has $2n+1$ divisors. So,

$2n+1=13\implies n=6 \implies N = p^n = p^6.$

Clearly, the smallest possible value of $N = p^6 = 2^6 = \boxed{64}$ .

The question should read $f(N)=13$ . If $a,b$ are positive integer solutions of $a^{-1}+b^{-1}=N^{-1}$ , then $a,b > N$ and $Nb+Na=ab$ , so $(a-N)(b-N)=N^2$ . Thus the positive integer solutions of the equation $a^{-1}+b^{-1}=N^{-1}$ are $a=N+\alpha,b=N+\beta$ where $\alpha,\beta$ are positive integer factors of $N^2$ and $\alpha\beta=N^2$ . Thus $f(N)=d(N^2)$ is the number of positive divisors of $N^2$ , which is always odd. For $N=p^n$ and $f(N)=13$ , we need $2n+1=d(p^{2n})=13$ , so $n=6$ . The smallest such $N$ is $N=2^6=64$ .

We expand to get that $\frac{a+b}{ab}=\frac{1}{x}$ , so $ab=x(a+b)$ . Thus $ab-x(a+b)=0$ and $ab-x(a+b)+x^2=x^2$ . Simplifying, we get $(a-x)(b-x)=x^2$ . Note that $a-x$ and $b-x$ must be positive integers, otherwise one of $a$ or $b$ would not be a positive integer. We first choose $a-x$ as a positive factor of $x^2$ then $b-x$ would follow. Thus $f(N)$ is the number of ways to choose a factor of $x^2$ , which means that $x^2$ must have 13 positive integer factors.

Thus $x^2$ is a 12th power of a prime, thus its minimum is $2^{12}=4096$ . Hence the smallest possible value of $N$ is $\sqrt{4096}=\boxed{64}$ .

Let $N = p^k,$ where $p$ is a prime and $k$ is a positive integer. Then we have $\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{p^k}.$ Clearing out the fractions, we get $ap^k+bp^k = ab.$ Manipulating this equation allows us to obtain the factorization $p^{2k} = (a-p^k)(b-p^k).$ Factoring the left-hand side as $p^{2k} = (p^i)(p^{2k-i})$ for some integer $i$ gives the solution $(a,b) = (p^i + p^k), (p^{2k-i} + p^k).$ Because $a$ and $b$ are integers, we must have $0 \le i \le 2k.$ In addition, we can't factorize $p^{2k} = (-p^i)(-p^{2k-i})$ because then $a$ and $b$ would not both be positive; so, these are the only solutions.

For each integer $i$ with $0 \le i \le 2k,$ there is exactly one solution to the original equation. Thus, there are $2k+1$ solutions to the original equation. Because we want $f(N) = 13,$ we solve the equation $2k+1 = 13$ to get $k=6.$ So, $N$ is in the form $p^6$ for some prime $p.$ To minimize $N,$ we take $p = 2,$ hence the smallest possible value of $N$ is $2^6 = \boxed{64}.$

We can manipulate the equation:

$\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{x}$

$\dfrac{a+b}{ab} = \dfrac{1}{x}$

$x(a+b) = ab$

$ab - x(a+b) + x^2 = x^2$

$ab - ax - bx + x^2 = x^2$

$(a-x)(b-x) = x^2$

Since $a,b > 0$ , then $a-x, b-x > -x$ . Suppose that $a-x \le 0$ , then $b-x \le 0$ (otherwise $(a-x)(b-x) \le 0$ while $x^2 > 0$ ). But then $(a-x)(b-x) = (x-a)(x-b) < x \cdot x = x^2$ , impossible. So $a-x > 0$ . Similarly, $b-x > 0$ .

Now, $a-x$ can be any positive factor of $x^2$ , and $b-x$ will simply follow suit ( $b-x = \dfrac{x^2}{a-x}$ ). Obviously as $a-x$ is a factor of $x^2$ , $\dfrac{x^2}{a-x}$ is a positive integer and hence a possible value for $b-x$ . On the other hand, when $a-x$ is not a factor of $x^2$ , $\dfrac{x^2}{a-x}$ is not an integer and hence $b-x$ will not be an integer, and hence $b$ will not be an integer, contradiction. So there are precisely as many solutions of $(a,b)$ as there are positive factors of $x^2$ . So $f(x)$ is precisely $\tau(x^2)$ where $\tau$ is the divisor function .

Now, let $N = p^n$ because $N$ is a prime power. Then $f(N) = f(p^n) = \tau(p^{2n}) = 2n+1$ , since there are $2n+1$ divisors of $p^{2n}$ (namely $p^0, p^1, \ldots, p^{2n}$ ). Since $f(N) = 13$ , we have $2n+1 = 13$ and so $n = 6$ .

Since $p^6$ is strictly increasing for positive $p$ , the smallest value of $N$ is achieved when $p$ is minimized. The smallest prime is $2$ , so $p=2$ gives the minimum value of $N = 2^6 = \boxed{64}$ .

We rewrite the equation:

$\frac1a + \frac1b = \frac1x \\ \iff \frac{a+b}{ab} = \frac1x \\ \iff ab = (a+b)x \\ \iff x^2 = ab - (a+b)x + x^2 = (a-x)(b-x)$

Therefore, the pair $(a,b)$ is a solution iff $(a-x)(b-x)$ is a factorization of $x^2$ with $a-x$ and $b-x$ both larger than $-x$ . This last condition rules out factorizations involving negative factors, since then one of them is necessarily at most $-x$ . All factorizations with positive factors however yield a valid solution.

It is specified that $x = p^n$ . The factorizations of $x^2 = p^{2n}$ are all of the form $p^k p^{2n-k}$ , with $0 \leq k \leq 2n$ , hence there are $2n+1$ factorizations and thus $2n+1$ solutions to the original equation. Therefore, $2n+1=13$ , and $n=6$ .

The smallest number $x$ of the form $p^6$ is trivially $x = 2^6 = \boxed{64}$

$\frac{1}{N}$ = $\frac{1}{a}$ + $\frac{1}{b}$

So, N is equal to:

N= $\frac{ab}{a+b}$

ab=aN + bN

ab - aN - bN=0

ab - aN - bN - N $^{2}$ =N $^{2}$

(a-N)(b-N)=N $^{2}$

$N^{2}$ is a perfect square and N is a prime power, therefore $N^{2}$ must have 13 divisors in order to f(N)=12, because each divisor correspond to one ordered pair(a,b), but N, because $N^{2}$ is a perfect square

$\frac{1}{a} + \frac{1}{b} = \frac{1}{N} \Rightarrow (a - N)(b - N) = N^{2}$ . Let $N = p^{k}$ . $(a - p^{k})(b - p^{k}) = p^{2k}$ have clearly $2k$ solutions. We want $12$ solutions, so we set $k = 6$ . The minimum value of $N$ is clearly $2^{6} = 64$ .

Let $N=p^n$ . Thus we examine the equation $\frac{1}{a} + \frac{1}{b} = \frac{1}{p^n}$ . Rearranging slightly, we have:

$\frac{1}{p^n} = \frac{a+b}{ab}$

$\therefore p^n = \frac{ab}{a+b}$

Let us say, without loss of generality, that $a \leq b$ . Since $p^n$ is an integer, $a+b$ must divide $ab$ . It follows that $b$ must be a multiple of $a$ and thus can be written as $ka$ , for some integer $k$ . Plugging in, we have:

$\frac{ab}{a+b} = \frac{ka^2}{a+ka} = \frac{ka}{1+k} = p^n$

$k$ is not a factor of $1+k$ , thus $1+k$ must divide $a$ . Since $p$ is prime, we have $k = p^c$ and $\frac{a}{1+k} = p^{n-c}$ , for some integer $c$ such that $0 \leq c \leq n$ . Solving for $a$ and $b$ , we have:

$a = (1+k)p^{n-c} = (1+p^c)p^{n-c}$

$b = ka = (1+p^c)p^n$

Plugging in valid values for $c$ ( $c$ is an integer such that $0 \leq c \leq n$ ) will always result in a pair of integers $(a, b)$ such that $\frac{1}{a} + \frac{1}{b} = \frac{1}{p^n}$ . Additionally, for every ordered pair $(a, b)$ obtained, we also have an additional pair $(b, a)$ , except for when $c = 0$ , because then $a = b$ . Since we want $f(N) = 13$ , we want $c$ to take on $\frac{13-1}{2} = 6$ values, not including $0$ . We thus we choose $n = 6$ .

The smallest prime power where $n = 6$ is $2^6 = 64$ .

Rearranging the expression we got: $b = \frac{ax}{a-x}$

So $a-x$ $|$ $ax$ , what implies that:

$\Rightarrow a-x$ $|$ $ax - x(a-x)$

$\Rightarrow a-x$ $|$ $x^2$

Therefore, we have as many $a$ 's satisfying the given conditions as divisors of $x^2$ . The smallest number that have exactly $13$ divisors and is a prime power is $2^{12}$ . Therefore, the number we're looking for is $\sqrt{2^{12}} = 2^6 = 64$ .

Clearly, $a,b > x$ . Multiply both sides of $\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{x}$ by $abx$ to get:

$bx+ax = ab$ , i.e. $0 = ab - ax - bx$ .

Using SFFT we can add $x^2$ to both sides to get:

$x^2 = ab - ax - bx + x^2 = (a-x)(b-x)$ .

Hence, $a-x$ and $b-x$ must be complementary positive factors of $x^2$ .

If $x^2$ has $m$ positive factors, $1 = f_1 < f_2 < \cdots < f_{m-1} < f_m = x^2$ , then:

$(a-x,b-x) = (f_k,f_{m-k})$ , i.e. $(a,b) = (x+f_k,x+f_{m-k})$ for $k \in \overline{1,m}$ .

There are $m$ such pairs $(a,b)$ . So, $f(x)$ is the number of positive factors of $x^2$ .

If $N = p^n$ , then $N^2 = p^{2n}$ which has $2n+1$ positive factors.

So, $f(N) = f(p^n) = 2n+1 = 13 \leadsto n = 6$ . So, we need $N = p^6$ .

The smallest prime is $p = 2$ . Hence, the smallest such $N$ is $2^6 = \boxed{64}$ .

We can try the prime power 2^6 or 64. (1/a) + (1/b) = 1/64. (a + b)/(ab) = 1/64. 64a + 64b = ab. a(b - 64) = 64b. From this equation we can find the a in term of b, namely: a = 64 + (64*64)/(b - 64). We can have 13 pairs of (a,b): b - 64 =1, 2,4,8,16,32,64,128,256,512,1024,2048,4096. So (a,b) = (4160,65),(2112,66),(1088,68),(576,72),(320,80),(192,96),(128,128),(96,192),(80,320),(72,576),(68,1088),(66,2112),(65,4160).

Rearranging gives:

- $b=x+\frac{x^{2}}{a-x}$

We hence have positive solutions corresponding to positive factors of x^2. If x is a prime power p^n then f(x)=2n. We therefore want n=6 and 64 is the smallest sixth power.

N.B. We presumably we cannot have a=b because otherwise f(x) would need to be odd. Two orders for distinct solutions and just one order for a=2x, b=2x.

We have $\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{N}$ . Clearing denominators we get $ab=Na+Nb\implies ab-Na-Nb=0\implies (a-N)(b-N)=N^2$ . Note that the number of factors of $N^2$ is also going to be the number of ordered pairs (a,b) satisfying the equation. But wait! There are an even number of solutions, and an odd number of factors of $N^2$ ! Where did we go wrong? Notice that the ordered pair (0,0) satisfies our equation after clearing denominators, but it does not satisfy our original equation. We had assumed that $a, b\ne 0$ but included it in our count. Therefore, we must subtract this case from our total count. So we need to find an $N$ such that the factors of $N^2$ minus one is equal to twelve. Therefore, the number of factors of $N^2$ is 13. We represent $N=p^n$ . So the number of factors of $p^{2n}$ is 13. Therefore $2n+1=13$ and $n=6$ . We want $N$ as small as possible, so we make $p=2$ . Therefore our final answer is $2^6=\boxed{64}$