- #1

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__not__be written on the form a = k

_{1}^2 or b = k

_{2}^2 where k

_{1}and k

_{2}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)

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- Thread starter johann1301h
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- #1

- 71

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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

TeethWhitener

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- #3

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Yes, i see. But k_{1}and k_{2}are rational (not necessarily integers), then it's trivial. Choose a = 1/16 and b = 9/16. Then you have r = 1.

- #4

- 16,458

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##(a;b;r) = (6,25;2,25;4)##

- #5

- 71

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Could you elaborate?##(a;b;r) = (6,25;2,25;4)##

- #6

TeethWhitener

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kYes, i see. But k1 and k2 are indeed integers.

- #7

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I understand that 1/4, 3/4, 1/16 and 9/16 are all not integers. But what I am wondering is if kk_{1}=1/4 and k_{2}=3/4, neither of which are integers.

- #8

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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!

- #9

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Yes, thats right.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

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http://math.stackexchange.com/questions/278935/can-a-finite-sum-of-square-roots-be-an-integer

(The first answer is for a more general case but is also more complicated).

- #11

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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.

- #12

TeethWhitener

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

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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