An Algebraic Problem

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saifmd
Posts:15
Joined:Wed Feb 24, 2021 12:41 am
An Algebraic Problem

Unread post by saifmd » Mon Mar 29, 2021 9:21 pm

If $a, b, c$ are natural numbers so that $\left(1 + \dfrac{1}{a}\right)\left(1 + \dfrac{1}{b}\right)\left(1 + \dfrac{1}{c}\right) = 2$, then figure out all the possible values of $a, b$ and $c$.
Last edited by tanmoy on Thu Apr 01, 2021 2:49 am, edited 1 time in total.

~Aurn0b~
Posts:46
Joined:Thu Dec 03, 2020 8:30 pm

Re: An Algebraic Problem

Unread post by ~Aurn0b~ » Mon Mar 29, 2021 10:49 pm

saifmd wrote:
Mon Mar 29, 2021 9:21 pm
If $(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})=2$, then figure out all the possible values of a, b and c.
are the a,b,c integers? or natural numbers?

saifmd
Posts:15
Joined:Wed Feb 24, 2021 12:41 am

Re: An Algebraic Problem

Unread post by saifmd » Tue Mar 30, 2021 10:01 am

~Aurn0b~ wrote:
Mon Mar 29, 2021 10:49 pm
saifmd wrote:
Mon Mar 29, 2021 9:21 pm
If $(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})=2$, then figure out all the possible values of a, b and c.
are the a,b,c integers? or natural numbers?
a, b, c are natural numbers.

~Aurn0b~
Posts:46
Joined:Thu Dec 03, 2020 8:30 pm

Re: An Algebraic Problem

Unread post by ~Aurn0b~ » Tue Mar 30, 2021 2:12 pm

saifmd wrote:
Mon Mar 29, 2021 9:21 pm
If $(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})=2$, then figure out all the possible values of a, b and c.
$\textbf{Solution}$
We have the equation, $(a+1)(b+1)(c+1)=2abc$
$\Rightarrow ab+bc+ca+a+b+c+1=abc$
$\Rightarrow 2(a+b+c)=abc-ab-bc-ca+a+b+c-1$
$\Rightarrow 2(a+b+c)=(a-1)(b-1)(c-1)$

Now, substituting $x=a-1,y=b-1,z=c-1$, we get, \[ 2(x+y+z+3)=xyz......(1)\]
All the $x,y,z$ are natural number, because if any one of $a,b,c$ is 1, suppose its c, we'll have $(a+1)(b+1)=ab$, which is not possible.
WLOG assume that $x\geq y\geq z$,
If $x=1,x=y=z=1$ which doesnt satisfy $(1)$
If $x=2$ then the highest value of RHS is 8, while the LHS is always greater than 8.
So, $x\geq 3\Rightarrow \frac{3}{x} \leq1$

we have, $xyz=2x+2y+2z+6\leq 6x+6$
$\Rightarrow yz-6\leq \frac{6}{x}\leq 2$
$\Rightarrow yz\leq 8$
$\Rightarrow 8\geq yz\geq z^2$
So the value of $z$ is either 1 0r 2.
  • Case 1: $z=1$
    \[ 2(x+y+4)=xy\]
    \[ (x-2)(y-2)=12\]
    So, $(x-2,y-2)=(12,1),(6,2),(4,3)$
    SO we get three solutions $(a,b,c)=(15,4,2),(9,5,2),(7,6,2)$
  • Case 2: $z=2$
    \[ 2(x+y+5)=2xy\]
    \[ (x-1)(y-1)=6\]
    So, $(x-1,y-1)=(6,1),(3,2)$
    So, we get 2 more solutions $(a,b,c)=(8,3,3),(5,4,3)$
Therefore all the solutions to the diophantine equation is
$(a,b,c)=(15,4,2),(9,5,2),(7,6,2),(8,3,3),(5,4,3)$
And all of its permutations.$\blacksquare$

saifmd
Posts:15
Joined:Wed Feb 24, 2021 12:41 am

Re: An Algebraic Problem

Unread post by saifmd » Tue Mar 30, 2021 4:37 pm

ভাই, এই অঙ্কটা যে বেসিক ও দক্ষতা দরকার পড়েছে৷ সেগুলো কীভাবে অর্জন করেছেন, একটু গাইডলাইন দিলে উপকৃত হতাম খুব। অনেক অনেক ধন্যবাদ।

~Aurn0b~
Posts:46
Joined:Thu Dec 03, 2020 8:30 pm

Re: An Algebraic Problem

Unread post by ~Aurn0b~ » Tue Mar 30, 2021 6:24 pm

saifmd wrote:
Tue Mar 30, 2021 4:37 pm
ভাই, এই অঙ্কটা যে বেসিক ও দক্ষতা দরকার পড়েছে৷ সেগুলো কীভাবে অর্জন করেছেন, একটু গাইডলাইন দিলে উপকৃত হতাম খুব। অনেক অনেক ধন্যবাদ।
The more you solve problems the better problems solver you'll become, so solve more and more problems.
AoPS forums helped me a lot for learning problem solving

Faiyaz-M-Nirob
Posts:1
Joined:Fri Feb 19, 2021 12:57 pm

Re: An Algebraic Problem

Unread post by Faiyaz-M-Nirob » Wed Mar 31, 2021 12:00 am

Full Solution of this problem: https://youtu.be/iYZFWq6sEsc

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