Russian 46 solution

Russian problems

Problem 46

Find all possible integer solutions for sqrt(x + sqrt(x ... (x + sqrt(x)) ... )) = y, where there are 1998 square roots.

Solution

Let s₁ = sqrt(x), s₂ = sqrt(x + s₁), s₃ = sqrt(x + s₂) and so on. So the equation given is y = s₁₉₉₈. We show first that all s_n must be integral for 1 <= n <= 1998. y is integral, so s₁₉₉₈ is integral. Now suppose s_n is integral. Then s_n-1 = s_n² - x is integral, proving the claim.

So in particular s₁ and s₂ are integers and s₂² = s₁² + s₁. But if s₁ > 0, then s₁² < s₁² + s₁ < (s₁ + 1)², which is impossible. Similarly s₁ < 0 is impossible. So the only possible solution is s₁ = 0 and hence x = 0 and y = 0.

Russian 41-50