Show that [√n + √(n + 1)] = [√(4n + 2)] for positive integers n.
Solution
We have ( √n + √(n + 1) )2 = 2n + 1 + 2 √(n2 + n). Now n2 < n2 + n < n2 + n + 1/4 = (n + 1/2)2, so √(4n+1) < √n + √(n+1) < √(4n+2). Hence [ √(4n+1) ] ≤ [ √n + √(n+1) ] ≤ [ √(4n+2) ]. But a square must be congruent to 0 or 1 mod 4, so [ √(4n+2) ] = [ √(4n+1) ]. Hence result.
√ John Scholes
jscholes@kalva.demon.co.uk
11 Mar 2002