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The Power(ful) Equations (Posted on 2007-12-22) Difficulty: 2 of 5
If x and y are positive integers, L: LCM of x and y, G: GCD of x and y, then solve the following equations for x and y values:
1) xy=LG, L>x.
2) x2+y2 = L2+G2, L>x>G.

See The Solution Submitted by Praneeth    
Rating: 4.0000 (2 votes)

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Solution LG problem! | Comment 3 of 4 |

I will start with the second equation cos is more easy to prove!

We have LG=xy (1) and so

(x+y)^2 = (L+G)^2

x+y=L+G (2)

so from (1) and (2) x=L,y=G or x=G, y=L but no one can be posible because

L>x>G so no solution for this equation.

About the second after use logarithm and use again LG=xy

we can write that

ln(x)/x = ln(L)/L

If we study the f(x) = ln(x)/x this function is increase on (1,e) and decrease after e.  Because L>x and the only integer positiv value between 1 and e is 2 this is the only solution we have.

x=2,y=4,L=4,G=2


  Posted by Chesca Ciprian on 2007-12-22 16:46:26
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