Prove or disprove the claim that for any positive integer m there is a positive integer n such that mn + 1 is a perfect square.

First, the claim can be expressed as:

\[ \forall{(m,n \in \mathbb{N})} \exists {(r \in \mathbb{Z})} [m \cdot n + 1 = r^2] \]

Observe that:

\[ \begin{aligned} m \cdot n + 1 & = r^2 \\ m \cdot n & = r^2 - 1 \\ m \cdot n & = (r + 1)(r - 1) \\ n & = \frac{(r + 1)(r - 1)}{m} \\ \end{aligned} \]

This means if the claim is true, m must divide either (r + 1) or (r - 1):

\[ \begin{aligned} \exists{(x \in \mathbb{N})} [(m \cdot x = r + 1) \lor (m \cdot x = r - 1)] \end{aligned} \]

Or,

\[ \begin{aligned} \exists{(x \in \mathbb{N})} [(r = m \cdot x - 1) \lor (r = m \cdot x + 1)] \end{aligned} \]

For example, when m = 1, the smallest value of r such that n > 0 would be when x = 3. So:

\[ \begin{aligned} r & = 1 \cdot 3 - 1 & = 2 \\ n & = \frac{(2 + 1)(2 - 1)}{1} & = 3 \\ m \cdot n + 1 & = 1 \cdot 3 + 1 & = 2^2 \end{aligned} \]

Given m >= 1, we can always pick a value x such that:

\[ \begin{aligned} r = m \cdot x - 1 >= 2 \end{aligned} \]

This is when:

\[ \begin{aligned} x >= \frac{3}{m} \end{aligned} \]

which means we can always find a value r (from m and x) and n (from r and m) that satisfies the claim. This completes the proof.

# posted by rot13(Unafba Pune) @ 4:22 PM