Determine the positive integer $x$ for which $\dfrac14-\dfrac{1}{x}=\dfrac16.$ Determine all pairs of positive integers $(a,b)$ for which $ab-b+a-1=4.$ Determine the number of pairs of positive integers $(y,z)$ for which $\dfrac{1}{y}-\dfrac{1}{z}=\dfrac{1}{12}.$ Prove that, for every prime number $p$, there are at least two pairs $(r,s)$ of positive integers for which $\dfrac{1}{r}-\dfrac{1}{s}=\dfrac{1}{p^2}.$
Problem
Source:
Tags: CSMC, CSMC 2018
enthusiast101
24.11.2018 21:08
a. $\dfrac14-\dfrac{1}{x}=\dfrac16 \implies \frac{-1}{x}=\frac{-1}{12} \implies \fbox{x=12}$.
enthusiast101
24.11.2018 21:10
c. It is just $(x,y)=(6,12), (x,y)=(-12,-6)$.
AlastorMoody
24.11.2018 21:11
for b, just factorise LHS, and the rest is trivial case bashing
minesnooper
24.11.2018 21:12
b. ab-b+a-1 = (b+1)(a-1) = 4 can just find pairs through factors of 4
blacksheep2003
24.11.2018 21:59
We can rewrite the LHS as $\frac{s-r}{rs}=\frac{1}{p^2}$, cross-multiplying gives us $sp^2-rp^2=rs$, and subtracting both sides gives us $rs+rp^2-sp^2=0$. Utilizing SFFT, we have $rs+rp^2-sp^2-p^4=-p^4$, and factoring the LHS gives us $(s+p^2)(r-p^2)=-p^4$, and we can have $s+p^2=p^3$ and $r-p^2=-p$ or $s+p^2=p^4$ and $r-p^2=-1$. Solving these gives us $(r,s)=(p^2-p,p^3-p^2)$ and $(r,s)=(p^2-1,p^4-p^2)$, respectively, which are positive for all primes $p$ $\blacksquare$
UnknownNingen
25.11.2018 03:14
For b. you can use simon's favorite factoring trick so and it will be easier if we move the terms in terms of degree: ab - b + a -1, (a-1)(b+1) = 4 4 is (4)(1) and (2)(2) and (1)(4) (no negatives because of restriction to positive integers) therefore (a,b) can be (5,0), (3,1), (2,3)
NotLeon
25.11.2018 06:58
enthusiast101 wrote: c. It is just $(x,y)=(6,12), (x,y)=(-12,-6)$. y and z are positive integers. The solution is similar to D. Answer should be
7
oralayhan
15.08.2021 11:27
for option $c$; There are $29$ integer solutions in total. $7$ of them are defined in positive integers.