1. **State the problem:** We want to find the number of Pythagorean triples $(a,b,c)$ with positive integers such that $a^2 + b^2 = c^2$, where $a < b$ and $c = b + 2$. 2. **Substitute $c = b + 2$ into the Pythagorean equation:** $$a^2 + b^2 = (b + 2)^2$$ 3. **Expand the right side:** $$a^2 + b^2 = b^2 + 4b + 4$$ 4. **Simplify by subtracting $b^2$ from both sides:** $$a^2 = 4b + 4$$ 5. **Rewrite as:** $$a^2 = 4(b + 1)$$ 6. **Since $a^2$ is divisible by 4, $a$ must be even. Let $a = 2k$ for some positive integer $k$:** $$ (2k)^2 = 4(b + 1) \\ 4k^2 = 4(b + 1) \\ k^2 = b + 1$$ 7. **Express $b$ in terms of $k$:** $$b = k^2 - 1$$ 8. **Recall $c = b + 2$, so:** $$c = k^2 - 1 + 2 = k^2 + 1$$ 9. **We have the triple:** $$(a,b,c) = (2k, k^2 - 1, k^2 + 1)$$ 10. **Check the condition $a < b$:** $$2k < k^2 - 1$$ 11. **Rewrite inequality:** $$k^2 - 2k - 1 > 0$$ 12. **Solve quadratic inequality:** The roots of $k^2 - 2k - 1 = 0$ are $$k = 1 \\pm \sqrt{2}$$ Since $k$ is positive integer, the inequality holds for $$k > 1 + \sqrt{2} \approx 2.414$$ So $k \geq 3$. 13. **Check for $k=3$:** $$(a,b,c) = (6, 8, 10)$$ which is a valid Pythagorean triple. 14. **Since $k$ can be any integer $\geq 3$, there are infinitely many such triples.** **Final answer:** The number of such triples is infinite.