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I Sqrt(a) + sqrt(b) = r, can r be whole?

  1. Jun 17, 2016 #1
    a and b are different natural numbers which can not be written on the form a = k1^2 or b = k2^2 where k1 and k2 are integers.

    r = √a + √b.

    can r be a natural number?

    (ive tried assuming r IS a natural number and then finding a contradiction, but without success)
     
  2. jcsd
  3. Jun 17, 2016 #2

    TeethWhitener

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    If we allow that k1 and k2 are rational (not necessarily integers), then it's trivial. Choose a = 1/16 and b = 9/16. Then you have r = 1.
     
  4. Jun 17, 2016 #3
    Yes, i see. But k1 and k2 are indeed integers.
     
  5. Jun 17, 2016 #4

    fresh_42

    Staff: Mentor

    ##(a;b;r) = (6,25;2,25;4)##
     
  6. Jun 17, 2016 #5
    Could you elaborate?
     
  7. Jun 17, 2016 #6

    TeethWhitener

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    k1=1/4 and k2=3/4, neither of which are integers.
     
  8. Jun 17, 2016 #7
    I understand that 1/4, 3/4, 1/16 and 9/16 are all not integers. But what I am wondering is if k1 and k2 are both integers, can r be a natural number?
     
  9. Jun 17, 2016 #8
    So to clarify since I see people keep giving fractions as proposed answers:

    Is there an r = √a + √b where a, b, and r are natural numbers and a and b are not squares? I like the question!
     
  10. Jun 17, 2016 #9
    Yes, thats right.
     
  11. Jun 17, 2016 #10
  12. Jun 17, 2016 #11

    fresh_42

    Staff: Mentor

    The answer is no, unless you allow negative square roots. ##a## and ##b## being non square integers mean they have at least one prime factor of odd degree. We may assume ##a## itself has only pairwise distinct primes (i.e. ##1## as their power).
    So we get ##a = (\sqrt{a})^2 = (r - \sqrt{b})^2 = r^2 + b - 2r\sqrt{b}## which can only hold for square numbers ##b##.
    But then ##a## is a natural number and the square root of single primes, which cannot be.
     
  13. Jun 17, 2016 #12

    TeethWhitener

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    Ah, I missed the word "natural" in the OP. Sorry about that.
     
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