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

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

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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2x^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

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

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

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

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

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

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

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

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

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

mathslover
 one year ago
Best ResponseYou've already chosen the best response.4if 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

mathslover
 one year ago
Best ResponseYou've already chosen the best response.4Now, 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)

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

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

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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2x^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

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

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

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

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

mathslover
 one year ago
Best 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.

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

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

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