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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 Solution, Part I | Comment 2 of 4 |
Let a = x/G, b = y/G

Then L = Gab.

Substituting into equation 1 yields

(Ga)^(Gb) = (Gab)^G

Simplifying,

(Ga)^b = Gab

(Ga)^(b-1) = b

The only integral solutions are is b =2, Ga = 2
      and b = 1.  But if b = 1, then Ga = x = L, which is ruled out by the problem.

If b = 2, then

G = 1, a = 2 does not lead to a solution, because this leads to
x = 2, y = 2, but the GCD of 2 and 2 is 2, not 1

So the only solution is G = 2, a = 1, b = 2
Then x = 2, y = 4, L = 4, G = 2.






  Posted by Steve Herman on 2007-12-22 15:18:13
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