Kvant Math Problem 906
Consider the equation
Verified: no
Verdicts: UNKNOWN + PASS
Solve time: 27m41s
Source on kvant.digital
Problem
- Prove that for every positive integer $a$, the equation $$\dfrac1x+\dfrac1y=\dfrac1a$$ has at least three solutions in positive integers $x$ and $y$.
- Find the number of positive integer solutions of this equation for $a=1985$.
M. V. Slavinskyi
Exploration
Consider the equation
$\frac1x+\frac1y=\frac1a,$
where $a,x,y$ are positive integers.
Multiplying by $axy$ gives
$a(x+y)=xy.$
Rearranging and adding $a^2$ to both sides yields
$xy-a(x+y)+a^2=a^2,$
hence
$(x-a)(y-a)=a^2.$
The problem is thus transformed into counting factor pairs of $a^2$.
Before proceeding, one must address the statement of part (1). For $a=1$,
$(x-1)(y-1)=1,$
so the only solution is
$(x,y)=(2,2).$
Thus the claim that there are at least three solutions for every positive integer $a$ is false as written. The correct statement is that the claim holds for every integer $a\ge 2$.
Problem Understanding
The transformation
$(x-a)(y-a)=a^2$
suggests a correspondence between solutions and positive divisors of $a^2$. Determining that correspondence gives both a lower bound for the number of solutions and an exact count when $a=1985$.
Since the original statement is false for $a=1$, the solution must first identify that exception and then prove the asserted property for all $a\ge2$.
Proof Architecture
The first step is to show that every positive integer solution of
$\frac1x+\frac1y=\frac1a$
corresponds to a factorization of $a^2$.
The second step is to prove a bijection between positive divisors of $a^2$ and ordered solution pairs $(x,y)$.
The third step is to deduce that the number of solutions equals $\tau(a^2)$, where $\tau$ denotes the divisor-counting function.
The final step is to apply this formula to $a=1985$.
Solution
Starting from
$\frac1x+\frac1y=\frac1a,$
multiplication by $axy$ gives
$a(x+y)=xy.$
Hence
$xy-a(x+y)=0,$
and after adding $a^2$,
(x-a)(y-a)=a^2. \tag{1}
Conversely, expanding (1) gives
$xy-a(x+y)+a^2=a^2,$
which is equivalent to the original equation. Thus the two equations are equivalent.
Let $(x,y)$ be a positive integer solution. Since $a^2>0$, equation (1) shows that $x-a$ and $y-a$ have the same sign.
They cannot both be negative. Indeed, if $x<a$ and $y<a$, then
$\frac1x>\frac1a,\qquad \frac1y>\frac1a,$
so
$\frac1x+\frac1y>\frac2a>\frac1a,$
contradicting the equation.
Hence
$x-a>0,\qquad y-a>0.$
Let
$d=x-a.$
Then $d$ is a positive divisor of $a^2$, and from (1)
$y-a=\frac{a^2}{d}.$
Therefore every solution has the form
x=a+d,\qquad y=a+\frac{a^2}{d}, \tag{2}
where $d\mid a^2$.
Conversely, if $d$ is any positive divisor of $a^2$, define $(x,y)$ by (2). Then
=d\cdot\frac{a^2}{d} =a^2,$$so $(x,y)$ satisfies (1) and hence the original equation. Thus there is a bijection between positive divisors of $a^2$ and ordered positive integer solutions $(x,y)$. Consequently, the number of solutions is$$N(a)=\tau(a^2).$$For $a\ge2$, the divisors$$1,\quad a,\quad a^2$$of $a^2$ are distinct. Hence$$\tau(a^2)\ge3,$$and therefore$$N(a)\ge3.$$This proves that the equation has at least three positive integer solutions for every integer $a\ge2$. For $a=1$,$$(x-1)(y-1)=1,$$which gives the unique solution$$(x,y)=(2,2).$$Accordingly, the statement of part (1) is false if $a=1$ is included; the correct version is obtained by replacing “every positive integer $a$” with “every integer $a\ge2$”. For part (2),$$1985=5\cdot397.$$Since $\sqrt{397}<20$ and $397$ is not divisible by any of the primes$$2,3,5,7,11,13,17,19,$$the number $397$ is prime. Therefore$$1985^2=5^2\cdot397^2.$$Using the divisor-counting formula,$$\tau(1985^2)=(2+1)(2+1)=9.$$Hence the number of positive integer solutions is$$\boxed{9}.$$## Verification of Key Steps The equivalence between the original equation and$$(x-a)(y-a)=a^2$$is exact. Every solution yields a positive divisor $d=x-a$ of $a^2$, and every positive divisor of $a^2$ produces a solution through$$x=a+d,\qquad y=a+\frac{a^2}{d}.$$The correspondence is bijective, so the number of ordered solutions is precisely $\tau(a^2)$. For $a=1985$,$$1985^2=5^2\cdot397^2,$$and therefore$$\tau(1985^2)=9.$$## Alternative Approaches Introducing new variables$$u=x-a,\qquad v=y-a$$transforms the equation directly into$$uv=a^2.$$Counting positive factor pairs $(u,v)$ of $a^2$ immediately gives the same formula$$N(a)=\tau(a^2).$$This approach leads to the same conclusion: for every $a\ge2$ there are at least three solutions, while for $a=1985$ there are exactly$$\boxed{9}$$positive integer solutions.