How long is this street?

If we choose house number $a$ , then we require $\begin{array}{rcl} \tfrac14a^2(a-1)^2 & = & \tfrac14n^2(n+1)^2 - \tfrac14a^2(a+1)^2 \\ 2a^2(a^2+1) & = & n^2(n+1)^2 \; = \; \big[n(n+1)\big]^2 \end{array}$ Thus we deduce that $2(a^2+1)$ must be a perfect square. Since $2(a^2+1)$ is even, we deduce that it must be equal to the square $(2b)^2$ of an even number, and hence $a^2 + 1 = 2b^2$ , with $n(n+1) = 2ab$ .

Thus we are looking for positive integer solutions of the equations $a^2 - 2b^2 \; = \; -1 \qquad n(n+1) \; = \; 2ab$ for $n < 10^7$ . The positive integer solutions of $a^2 - 2b^2 = -1$ are given by the formula $a_k + b_k\sqrt{2} \; = \; \big(\sqrt{2}+1\big)^{2k+1} \qquad k \in \mathbb{N} \cup \{0\} \;.$ and the matching value of $n$ is the positive solution $n_k$ of the equation $n(n+1) = 2a_kb_k$ .

The resulting value of $n_k$ is less than $10^7$ just when $0 \le k \le 8$ , and of these $9$ cases $n_k$ is an integer only when $k=0$ , in which case $n_0=1$ .

The only value of $n$ that works is $n=\boxed{1}$ .