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

  • #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)
 

Answers and Replies

  • #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.
 
  • #3
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.
Yes, i see. But k1 and k2 are indeed integers.
 
  • #6
TeethWhitener
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Yes, i see. But k1 and k2 are indeed integers.
k1=1/4 and k2=3/4, neither of which are integers.
 
  • #7
k1=1/4 and k2=3/4, neither of which are integers.
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?
 
  • #8
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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!
 
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  • #9
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!
Yes, thats right.
 
  • #11
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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.
 
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  • #12
TeethWhitener
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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!
Ah, I missed the word "natural" in the OP. Sorry about that.
 

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