Question

A nice problem :)

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

Answer 1

im stuck at y/x = (x,y)

am i in right direction ?

Answer 2

what did u do, plz show ur work briefly...and i dont know the answer :)

Answer 3

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

Answer 4

me too

Answer 5

that implies if y|x, then x,y is the solution of equation.
all multiplies of of numbers is the solution of it

Answer 6

emm 2|6 but (2,6) is not a solution...

Answer 7

like take y=100, x=50. that gives gcd = 2 which is not a solution

Answer 8

wopps!! sorry wrong conclusion

Answer 9

:D np man

Answer 10

Got it ! (2, 4) is a solution

all (x, x^2) pairs less than 100 will work

Answer 11

and only these will work. so total 10 solutions

Answer 12

thats right :) 10 solutions... how did u do it?

Answer 13

They all will be of the form : (x, x^2)

Answer 14

(1,1) , (2,4) ... (10,100)

Answer 15

So, 10 solutions.

Answer 16

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

Answer 17

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)

Answer 18

hey i'll come back to this later :)

Answer 19

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)}\)

Answer 20

Ok mukushla, will wait for your response. Sorry for interrupting your solution @ganeshie8 , but is it same to mine?

Answer 21

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

Answer 22

Also right, nice!

Answer 23

Excellent work @mathslover

Answer 24

thanks

Answer 25

mathslover...how u come up with x^2=lambda and y^2=lambda^2 ?

Answer 26

Actually it involves a large algebra.

Answer 27

Wait please.

Answer 28

looks like you guys nailed it.

Answer 29

\(\mathsf{y^2 = \lambda x \\
x^2 = \lambda y \\
Therefore, ~ \cfrac{y^4}{\lambda^2 } = \mu y \\
\implies y^3 = \mu \lambda ^2 \\
Therefore, ~ x^6 = \lambda^3 y^3 \\
\implies x^6 = \lambda^2 \mu ^4 \\
\implies x^3 = \lambda \mu^2 \\
Now, ~ I ~ had ~ already ~ calculated ~ two ~ solutions ~ (1,1) ~ and ~ (2,4) . \\
Putting ~ them ~ in ~ the ~ equations ~ I ~ got: \mu = 1 \\
And ~ thus ~ the ~ answer . }\)

I agree that it looks absurd but it didn't strike the way ganeshi8 did.

Answer 30

@ganeshie8 and @mathslover :)

Answer 31

thanks muku for the beautiful problem :)

Answer 32

Yep. It was really a nice problem mukushla.