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Make it solvable (Posted on 2005-05-05) Difficulty: 4 of 5
a, b, and x are positive integers such that

sqrt(a) + sqrt(b) = sqrt(x)

How many possible values of x less than or equal to 1000 are there?

  Submitted by Jer    
Rating: 3.6667 (3 votes)
Solution: (Hide)
392

This is solvable if and only if x has a perfect square factor. proof:
x = a + b + sqrt(ab) so ab must be square
if a has an odd powered factor then b has the same odd powered factor.
sqrt(a) simplifies to a number of form d*sqrt(c) implied sqrt(b) simplifies to e*sqrt(c)
therefore
sqrt(x) = d*sqrt(c) + e*sqrt(c) = (d+e)*sqrt(c)
x = (d+e)^2 * c
so x has a perfect square factor.

Counting the number of integers which have a square factor takes awhile but is trivial.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionComputer solutionLarry2023-03-25 09:25:31
Some ThoughtsPuzzle Thoughts K Sengupta2023-02-12 07:51:01
SolutionAnswer?Justin2005-05-15 05:46:54
SolutionanswerCarey2005-05-11 22:52:05
Solutioneasily solvedSamantha2005-05-07 19:41:46
SolutionRounding errorCharlie2005-05-06 13:47:18
re(2): Independent SolutionTristan2005-05-06 05:53:45
Some Thoughtsre: Independent SolutionJohn Reid2005-05-06 02:12:12
SolutionIndependent SolutionTristan2005-05-05 23:35:58
SolutionThe correct(?) solutionJohn Reid2005-05-05 22:04:25
re(5): First thoughts... ...(spoiler?)Steve Herman2005-05-05 20:15:00
Some ThoughtsGeneral solutionOld Original Oskar!2005-05-05 19:56:06
re(4): First thoughts... ...(spoiler?)Charlie2005-05-05 19:55:13
re(4): First thoughts... ...(spoiler?)Charlie2005-05-05 19:51:17
Solutionre(3): First thoughts... ...(spoiler?)Charlie2005-05-05 19:47:04
re(2): First thoughts... ...(spoiler?)Erik O.2005-05-05 19:18:56
re: First thoughts... ...(spoiler?)Steve Herman2005-05-05 19:02:03
Hints/TipsFirst thoughts... ...(spoiler?)Erik O.2005-05-05 18:51:48
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