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The root of the problem

Yesterday, while discussing various proofs and problems, my friend Steve posted this and said it was his favourite problem of the last few years.


He laid it down as a challenge for me and the others involved in the discussion to solve.

The question:

Find all solutions in non negative integers a,b to (a)^1/2 + (b)^1/2 = (2009)^1/2

I was out when I saw the question so couldn’t crack on with it straight away, but I did throw a few ideas around in my head. My first thought was “one equation, two unknowns, Hmmmm, I could square it, rearrange the original and sub it it to eliminate, actually no. I would just be proving a=a, b=b, and 2009=2009 then, so that’s a bit silly.” It always amuses me when I, or someone else, manages to prove something like this. My Year 13s still laugh about the time one of them started and finished with F = ma after two pages of working!

My next thought was to complete a prime factorisation on 2009. As it happens I didn’t need to my brother posted “I’ve simplified root 2009 to 7root41” which was what my prime factorisation was for. I realise now that I probably should have seen the answer and this point, but I didn’t.

I thought it would be a good idea to rearrange the equation to find a:


I replaced root 2009 with 7 root 41 and then it was clear: for a to be an integer, b has to be a multiple of 41. This is because any integer b would make the first and last terms integers, but only multiples of 41 will make the middle term an integer. The term is 14x(b)^1/2(41)^1/2 and you need a root 41 to rationalise the one that’s there. Moreover, b must be 41 multiplied by a square number, as any other factors of b would leave another surd.

From there it was easy to spot the answers.


The root a term plus the root b term must equal 7 root 41, so if root a = root 41 then root b = 6 root 41. Likewise if root a = 2 root 41 then root b = 5 root 41. In general if root a = n root 41 then root b = (7-n) root 41 (n is a natural number less than 7). Eliminate your roots and you get {a,b} = {41n^2, 41(7-n)^2} n € N, n < 7.


A truly lovely problem, with a thoroughly beautiful solution.

Nb I apologise for my inability to code the correct maths symbols etc.

  1. April 15, 2014 at 7:14 pm

    I got pretty much the same answer (although I’m not sure why you’ve listed it in terms of n rather than written the list of 6 possibilities) but I used the “fact” that if you simplify all your surds then all the terms of the equation must be multiples of the same surd (in this case root 41). However, I’m not actually sure how I knew this to be the case. Any idea what theorem or result I’m using? Why is it not possible to add different (simplified) surds and get a multiple of a different (simplified) surd?

    • April 15, 2014 at 7:59 pm

      Yes, that was what I realised at the end, and what I meant when I said I should have got it earlier! Not sure what the Theorem is that states that, it is due to the primarily of the number. I know exactly why it is, but I’m really struggling to know how to explain it! I will give it more thought and try and get back to you with an explanation!

      Re expressing in n, I don’t know why I didn’t just list them! Possibly I was “showing off”, or perhaps I’m just used to trying to fit things into 140 characters!

    • April 16, 2014 at 9:05 am

      When you simplify a surd to its simplest form you have the lowest number which is relatively prime to all the squares. If you try a prime factorisation you get the integer part expressed as a product if two primes but you are unable to break it down and further. When you add numbers that have a common factor your answer will always have that common factor. 3x + 6x =9x all three terms have common factors of 3 and x. It is this rule that we are using here.

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