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mukushla
Group Title
A nice problem :)
How many solutions are there for the equation\[x^2+y^2=xy(x,y)+[x,y]\]where\[(x,y)=\gcd(x,y)\]\[[x,y]=\text{lcm}(x,y)\]\[x,y \in \mathbb{N}\]\[x\le y \le100\]
 one year ago
 one year ago
mukushla Group Title
A nice problem :) How many solutions are there for the equation\[x^2+y^2=xy(x,y)+[x,y]\]where\[(x,y)=\gcd(x,y)\]\[[x,y]=\text{lcm}(x,y)\]\[x,y \in \mathbb{N}\]\[x\le y \le100\]
 one year ago
 one year ago

This Question is Closed

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
im stuck at y/x = (x,y) am i in right direction ?
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
what did u do, plz show ur work briefly...and i dont know the answer :)
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
x^2 + y^2 = xy(x,y) + [x,y] x^2 + y^2 = xy(x,y) + xy/(x,y) (x^2 + y^2)/xy = (x,y) + 1/(x,y) x/y + y/x = (x,y) + 1/(x,y) ... not sure how to conclude
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
that implies if yx, then x,y is the solution of equation. all multiplies of of numbers is the solution of it
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
emm 26 but (2,6) is not a solution...
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
like take y=100, x=50. that gives gcd = 2 which is not a solution
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
wopps!! sorry wrong conclusion
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
:D np man
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
Got it ! (2, 4) is a solution all (x, x^2) pairs less than 100 will work
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
and only these will work. so total 10 solutions
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
thats right :) 10 solutions... how did u do it?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.4
They all will be of the form : (x, x^2)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.4
(1,1) , (2,4) ... (10,100)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.4
So, 10 solutions.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.4
if we iobserve carefully then we will see that RHS is a multiple of both x and y (separately). So, LHS must also be a multiple f x and y
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.4
Now, x(x^2 + y^2) => x  y^2 Similarly, yx^2 Let, x y^2 => y^2 = lambda x and x^2 = mu y Thus, solving these two I get x^2 = lambda y^2 = lambda^2 Thus, (x,x^2) is the general solution, As, y should be between 0 and 100. So, x can range from 0 to 10. So here is the ans : (1,1),(2,4)...(10,100)
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
hey i'll come back to this later :)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.4
Now, \(\mathsf{x(x^2 + y^2)\\ \implies x  y^2 \\ Similarly, \\ yx^2 \\ Let, \\ x y^2 \\ \implies y^2 = \lambda x \\ and\\ x^2 = \mu y \\ Thus~ , ~ solving ~ these ~two ~ I ~get \\ x^2 = \lambda\\ y^2 = \lambda^2\\ Thus,~ \\ (x,x^2) ~is~ the ~general~ solution,\\ As,~ y ~ should ~be ~between~ 0 ~and ~100.\\ So,~ x~ can ~range ~from~ 0 ~to ~10.\\ So ~here ~is ~the ~ans ~: ~~(1,1),(2,4)...(10,100)}\)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.4
Ok mukushla, will wait for your response. Sorry for interrupting your solution @ganeshie8 , but is it same to mine?
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
x^2 + y^2 = xy(x,y) + [x,y] x^2 + y^2 = xy(x,y) + xy/(x,y) (x^2 + y^2)/xy = (x,y) + 1/(x,y) x/y + y/x = (x,y) + 1/(x,y) y/x = (x,y) y needs to be multiple of x, cuz (x,y) is a natural number. that gives gcd = x kx/x = x k = x y = x^2
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.4
Also right, nice!
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
Excellent work @mathslover
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
mathslover...how u come up with x^2=lambda and y^2=lambda^2 ?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.4
Actually it involves a large algebra.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.4
Wait please.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
looks like you guys nailed it.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.4
\(\mathsf{y^2 = \lambda x \\ x^2 = \lambda y \\ Therefore, ~ \cfrac{y^4}{\lambda^2 } = \mu y \\ \implies y^3 = \mu \lambda ^2 \\ Therefore, ~ x^6 = \lambda^3 y^3 \\ \implies x^6 = \lambda^2 \mu ^4 \\ \implies x^3 = \lambda \mu^2 \\ Now, ~ I ~ had ~ already ~ calculated ~ two ~ solutions ~ (1,1) ~ and ~ (2,4) . \\ Putting ~ them ~ in ~ the ~ equations ~ I ~ got: \mu = 1 \\ And ~ thus ~ the ~ answer . }\) I agree that it looks absurd but it didn't strike the way ganeshi8 did.
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
@ganeshie8 and @mathslover :)
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.2
thanks muku for the beautiful problem :)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.4
Yep. It was really a nice problem mukushla.
 one year ago
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