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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\[(x,y)=\gcd(x,y)\]\[[x,y]=\text{lcm}(x,y)\]\[x,y \in \mathbb{N}\]\[x\le y \le100\]

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im stuck at y/x = (x,y) am i in right direction ?
what did u do, plz show ur work briefly...and i dont know the answer :)
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

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Other answers:

me too
that implies if y|x, then x,y is the solution of equation. all multiplies of of numbers is the solution of it
emm 2|6 but (2,6) is not a solution...
like take y=100, x=50. that gives gcd = 2 which is not a solution
wopps!! sorry wrong conclusion
:D np man
Got it ! (2, 4) is a solution all (x, x^2) pairs less than 100 will work
and only these will work. so total 10 solutions
thats right :) 10 solutions... how did u do it?
They all will be of the form : (x, x^2)
(1,1) , (2,4) ... (10,100)
So, 10 solutions.
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
Now, x|(x^2 + y^2) => x | y^2 Similarly, y|x^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)
hey i'll come back to this later :)
Now, \(\mathsf{x|(x^2 + y^2)\\ \implies x | y^2 \\ Similarly, \\ y|x^2 \\ Let, \\ x| y^2 \\ \implies y^2 = \lambda x \\ and\\ x^2 = \mu y \\ Thus~ , ~ solving ~ these ~two ~ I ~get \\ x^2 = \lambda\\ y^2 = \lambda^2\\ Thus,~ \\ (x,x^2) ~is~ the ~general~ solution,\\ As,~ y ~ should ~be ~between~ 0 ~and ~100.\\ So,~ x~ can ~range ~from~ 0 ~to ~10.\\ So ~here ~is ~the ~ans ~: ~~(1,1),(2,4)...(10,100)}\)
Ok mukushla, will wait for your response. Sorry for interrupting your solution @ganeshie8 , but is it same to mine?
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
Also right, nice!
Excellent work @mathslover
thanks
mathslover...how u come up with x^2=lambda and y^2=lambda^2 ?
Actually it involves a large algebra.
Wait please.
looks like you guys nailed it.
\(\mathsf{y^2 = \lambda x \\ x^2 = \lambda y \\ Therefore, ~ \cfrac{y^4}{\lambda^2 } = \mu y \\ \implies y^3 = \mu \lambda ^2 \\ Therefore, ~ x^6 = \lambda^3 y^3 \\ \implies x^6 = \lambda^2 \mu ^4 \\ \implies x^3 = \lambda \mu^2 \\ Now, ~ I ~ had ~ already ~ calculated ~ two ~ solutions ~ (1,1) ~ and ~ (2,4) . \\ Putting ~ them ~ in ~ the ~ equations ~ I ~ got: \mu = 1 \\ And ~ thus ~ the ~ answer . }\) I agree that it looks absurd but it didn't strike the way ganeshi8 did.
thanks muku for the beautiful problem :)
Yep. It was really a nice problem mukushla.

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