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

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ganeshie8Best ResponseYou've already chosen the best response.2
im stuck at y/x = (x,y) am i in right direction ?
 10 months ago

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

ganeshie8Best 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
 10 months ago

experimentXBest 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
 10 months ago

mukushlaBest ResponseYou've already chosen the best response.0
emm 26 but (2,6) is not a solution...
 10 months ago

ganeshie8Best ResponseYou've already chosen the best response.2
like take y=100, x=50. that gives gcd = 2 which is not a solution
 10 months ago

experimentXBest ResponseYou've already chosen the best response.0
wopps!! sorry wrong conclusion
 10 months ago

ganeshie8Best 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
 10 months ago

ganeshie8Best ResponseYou've already chosen the best response.2
and only these will work. so total 10 solutions
 10 months ago

mukushlaBest ResponseYou've already chosen the best response.0
thats right :) 10 solutions... how did u do it?
 10 months ago

mathsloverBest ResponseYou've already chosen the best response.4
They all will be of the form : (x, x^2)
 10 months ago

mathsloverBest ResponseYou've already chosen the best response.4
(1,1) , (2,4) ... (10,100)
 10 months ago

mathsloverBest 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
 10 months ago

mathsloverBest 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)
 10 months ago

mukushlaBest ResponseYou've already chosen the best response.0
hey i'll come back to this later :)
 10 months ago

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

mathsloverBest 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?
 10 months ago

ganeshie8Best 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
 10 months ago

ganeshie8Best ResponseYou've already chosen the best response.2
Excellent work @mathslover
 10 months ago

mukushlaBest ResponseYou've already chosen the best response.0
mathslover...how u come up with x^2=lambda and y^2=lambda^2 ?
 10 months ago

mathsloverBest ResponseYou've already chosen the best response.4
Actually it involves a large algebra.
 10 months ago

experimentXBest ResponseYou've already chosen the best response.0
looks like you guys nailed it.
 10 months ago

mathsloverBest 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.
 10 months ago

mukushlaBest ResponseYou've already chosen the best response.0
@ganeshie8 and @mathslover :)
 10 months ago

ganeshie8Best ResponseYou've already chosen the best response.2
thanks muku for the beautiful problem :)
 10 months ago

mathsloverBest ResponseYou've already chosen the best response.4
Yep. It was really a nice problem mukushla.
 10 months ago
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