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

[johann1301h]

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)

 

[TeethWhitener]

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.

 

[johann1301h]

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.

 

[fresh_42]

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

 

[johann1301h]

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

Could you elaborate?

 

[TeethWhitener]

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

k₁=1/4 and k₂=3/4, neither of which are integers.

 

[johann1301h]

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?

 

[DocZaius]

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!

 

[johann1301h]

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, that's right.

 

[DocZaius]

The second answer here proves a and b must be squares:
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).

 

[fresh_42]

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.

 

[TeethWhitener]

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.