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

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
Gold Member
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.

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
Mentor
2021 Award
##(a;b;r) = (6,25;2,25;4)##

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

TeethWhitener
Gold Member
Yes, i see. But k1 and k2 are indeed integers.
k1=1/4 and k2=3/4, neither of which are integers.

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?

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!

• mfb
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.

fresh_42
Mentor
2021 Award
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.

• ProfuselyQuarky and micromass
TeethWhitener