Solution
If n+1 is a square then we only need to show that \sqrt{n} is irrational, which amounts to showing that it's not a square. This follows the fact that if n+1 and n are both squares then n=0, contradicting problem condition.
If n+1 is not a square, then it has prime factors with odd powers in n. Let k be the largest of such factor and a is its power. Meaning, k^a | n+1 but k^{a+1} does not divide n+1.
Now suppose there exist p,q relatively prime such that \sqrt{\frac{n}{n+1}}=\frac{p}{q} (n+1)p^2 = nq^2 Because k^a | LHS then k^a | RHS. If k^a | n then k | (n+1)-n = 1, a contradiction (because k is a prime).
And because k is prime then k^a | q^2 which means k^{a+1} | q^2 (because a is odd). This means that k divides p, contradicting our assumption that p,q are relatively prime.
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