None I suppose? Each of 2005/(x+y), 2005/(y+z), 2005/(x+z) must be an integer. Since 2005 is odd, x+y, y+z, x+z must be odd, so their sum 2(x+y+z) should be odd, contradiction.

Well, maybe I am wrong but you used the following statment : If $\sqrt{a} +\sqrt{b} +\sqrt{c}$ is an integer with $a,b,c \in \mathbb Q^{+}$ then each of $a,b,c$ must be an integer. This is false, because for example : Take $a=\frac{9}{4}$ , $b=\frac{25}{4}$ and $c=\frac{36}{4}$ We have $\sqrt{a} +\sqrt{b} +\sqrt{c}=7$ is an integer , but $a$ and $b$ aren't integers.

Simple. one can easily show that for the sum of the three square roots to be an integer, each of the terms should be a perfect square. So 2005/(x+y) = (p/q)^2 ,... Now 2005= 5(401) none of them being perfect squares. Hence x+y = 2005 n1^2 y+z=2005n2^2 z+x=2005 n3^2 and finally 1/n1 + 1/n2 +1/n3 is an integer. Note that the triple (n1,n2,n3) satisfying the above is a solution for (x,y,z) iff n1+n2+n3 be an even number. checking the cases we see easily that (n1,n2,n3) = (2,2,4),(2,4,2),(4,2,2) -Ali

n1,n2,n3 should be (4,4,2),... not (2,2,4)

what do u think of : sqrt((2-sqrt(3))^2)+sqrt((2+sqrt(3)^2))+sqrt(4)=8 ?? and u also forget to check in the expression the values u found !!!

here is what i think of your example: [2-sqrt(3)]^2 = 2005 / (x+y) is impossible!! Since each term is a rational it should be the perfect square of the rational. And i don't see exactly what i need to check? Anyways, i have to go now!

I think there is one triple (14*2005, 2*2005, 2*2005). Is it true?

Yes, it is BTW: I've seen some theorems about sums of square roots being an integer, but don't know where to find it now... Do you know?

ali wrote: one can easily show that for the sum of the three square roots to be an integer -Ali i cannot show

I didn't understand exactly what Ali meant, but here is the key idea [I just hope it's not the same thing what Ali said]: If $a,b,c\in\mathbb{Q}$ and $\sqrt{a}+\sqrt{b}+\sqrt{c}\in\mathbb{Q}$ then $\sqrt{a},\sqrt{b},\sqrt{c}\in\mathbb{Q}$. Proof: Because $s=\sqrt{a}+\sqrt{b}+\sqrt{c}\in\mathbb{Q}$, by squaring ,subtracting $a+b+c\in\mathbb{Q}$ and dividing by 2, we get $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\in\mathbb{Q}$. Let $\sqrt{abc}=p$. Repeating again the operation we get $\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})=s\sqrt{p}\in\mathbb{Q}$, that is $\sqrt{abc}\in\mathbb{Q}$. Now, $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=\sqrt{a}(\sqrt{b}+\sqrt{c})+\frac{\sqrt{abc}}{\sqrt{a}}=\sqrt{a}(s-\sqrt{a})+\frac{\sqrt{p}}{\sqrt{a}}\in\mathbb{Q}$. It results $\sqrt{a}\left(s+\frac{\sqrt{p}}{a}\right)-a\in\mathbb{Q}$. It results $\sqrt{a}\in\mathbb{Q}$. The same way we get that $\sqrt{b}$ and $\sqrt{c}$ are rational. ___ From now, the solution is easy, with the solution indeed being $(2005\cdot2,2005\cdot2,2005\cdot14)$ and its permutations.