, 2006

We use the known characterisation of primitive Pythagorean triples $(a, b, c)$: $a = 2pq$, $b = p^2 - q^2$, $c = p^2 + q^2$, where $p$ and $q$ are relatively prime integers, one even and $q < p$. The given condition entails $c^2 = a^2 + b^2 > 2 \times 2000^2$. Hence $2000\sqrt{2} < c = p^2 + q^2 < 3000$. Using the approximation $\sqrt{2} = 1.4142$, we get $2828 < p^2 + q^2 < 3000$. We also have $2000 < b = p^2 - q^2$. Adding these two, we obtain $4828 < 2p^2$, whence $50 \le p$. Again, we have $2000 < a = 2pq$. Subtracting this from $p^2+q^2 < 3000$, we get $(p-q)^2 < 1000$ giving $p-q \le 31$. Using $50 \le p$ and $p-q \le 31$, we obtain $19 \le q$ and hence $3000 > p^2+q^2 \ge p^2+19^2$. It follows that $p^2 < 2639$ and thus $p \le 51$. We obtain $2000/51 < 2pq/p = 2q$.

We thus obtain a better bound for $q$: $20 \le q$ and hence a better bound for $p$: $p^2 < 2639$ and hence $p \le 50$. We have seen earlier that $p \ge 50$, thus getting $p = 50$. Using $p^2 + q^2 < 3000$, we now get $q^2 < 500$. This implies that $q \le 22$. Thus $20 \le q \le 22$. However $q$ has to be odd since $p$ is even. We conclude that $q = 21$. We see that $a = 2100$, $b = 2059$ and $c = 2941$ give only one triple in the given range.