If $ n$ is prime, we need $ n|t$ as $ tn$ has to contain the prime factor $ n$ at least twice. But this gives $ t\geq n > \frac {n}{4}$. Additionally, $ x^{2}\equiv 1\mod p$ has only the solutions $ \pm 1\mod p$, thus for $ kn + 1$ to be a square $ a^{2}$, we need $ a\equiv\pm 1\mod p$. From $ a > 1$ we conclude $ a\geq n - 1$, yielding $ kn + 1\geq (n - 1)^{2}\iff k\geq n - 2$. But from $ n > 3$ we get $ n - 2 > \frac {n}{4}$, thus $ k > \frac {n}{4}$. If $ n$ is composite, we consider two cases: Case 1: $ n$ has at least two different prime divisors $ p\neq q$, let $ p^{a}$ be the highest power of $ p$ dividing $ n = p^{a}m$. We have $ m > 2$ (remember that $ m$ is odd) and $ \gcd(p^{a},m) = 1$, thus by the Chinese Remainder Theorem, we find an $ y$ such that $ y\equiv 1\mod p^{a}$ and $ y\equiv - 1\mod m$. We can assume that $ |y| < \frac {n}{2}$, too, giving $ y^{2} < \frac {n^{2}}{4}$. From $ y^{2}\equiv 1\mod p^{a}$ and $ \mod m$ we get $ y^{2}\equiv 1\mod n$, from $ y\not\equiv 1\mod m$ we get $ y\not\equiv 1\mod n$ and $ y\not\equiv - 1\mod p^{a}$ gives $ y\not\equiv - 1\mod n$. Thus if we set $ k = \frac {y^{2} - 1}{n}$, this is a positive (as $ y\neq\pm 1$) integer and it fulfills $ k = \frac {y^{2} - 1}{n} < \frac {y^{2}}{n} < \frac {\frac {n^{2}}{4}}{n} = \frac {n}{4}$. Case 2: $ n$ has only one prime divisor, thus $ n = p^{a}$ for a prime $ p$. If $ a$ is even, we can take $ t = 1 < \frac {n}{4}$ (as $ n\geq 5$ by assumption). If $ a$ is odd, it is $ \geq 3$ and we can take $ t = p < p\cdot\frac {p^{2}}{4} = \frac {p^{3}}{4}\leq\frac {n}{4}$. In both cases, at least one of $ k,t$ can be choosen such that $ kn + 1$ or $ tn$ is a square, but being lower than $ \frac {n}{4}$, in other words, we are done.