Three consecutive integers form the side lengths of a right angled triangle. How many distinct sets of such three consecutive positive integers are possible?

Introduction / Context:
This question combines number theory with the Pythagoras theorem. It asks how many triples of three consecutive positive integers can appear as the side lengths of a right angled triangle. Such triples, if they exist, must satisfy the Pythagorean condition for some ordering of the three numbers.

Given Data / Assumptions:

• The three side lengths are consecutive positive integers: n, n + 1 and n + 2 for some positive integer n.
• The triangle is right angled, so one side is the hypotenuse and the other two are legs.
• Because the numbers are consecutive, the largest side is n + 2.
• We are dealing only with positive integers as side lengths.

Concept / Approach:
For a right angled triangle with integer sides (a Pythagorean triple), we must have: (leg1)^2 + (leg2)^2 = (hypotenuse)^2 Given the sides are n, n + 1 and n + 2 with n + 2 as the largest, we need to test whether: n^2 + (n + 1)^2 = (n + 2)^2 Other arrangements where n or n + 1 is the hypotenuse are not possible, because the hypotenuse must be the largest side. Solving this equation will show how many integer values of n work.

Step-by-Step Solution:
Step 1: Let the three consecutive integers be n, n + 1 and n + 2. Step 2: Since n + 2 is the largest, consider it as the hypotenuse. Step 3: Apply Pythagoras theorem: n^2 + (n + 1)^2 = (n + 2)^2. Step 4: Expand the left side: n^2 + (n + 1)^2 = n^2 + (n^2 + 2n + 1) = 2n^2 + 2n + 1. Step 5: Expand the right side: (n + 2)^2 = n^2 + 4n + 4. Step 6: Set them equal: 2n^2 + 2n + 1 = n^2 + 4n + 4. Step 7: Rearrange: 2n^2 + 2n + 1 − n^2 − 4n − 4 = 0, which simplifies to n^2 − 2n − 3 = 0. Step 8: Factor: n^2 − 2n − 3 = (n − 3)(n + 1) = 0. Step 9: So n = 3 or n = −1. Only positive integer n is allowed, so n = 3. Step 10: The corresponding triple is (3, 4, 5), which is the famous Pythagorean triple. Step 11: Therefore, only one set of three consecutive positive integers works: 3, 4 and 5.

Verification / Alternative check:
Check the triple (3, 4, 5): 3^2 + 4^2 = 9 + 16 = 25 and 5^2 = 25, which satisfies the Pythagoras theorem exactly. For any larger n, the equation reduces to the same quadratic, which has no other positive integer solution. So no other consecutive positive integer triple can form a right angled triangle.

Why Other Options Are Wrong:
Only two or only three: These would require additional integer solutions to the quadratic equation, which do not exist.
Infinitely many: This is impossible because the equation reduces to a fixed quadratic with only two roots, of which only one is positive.
None: This would be true only if even 3, 4, 5 failed the Pythagorean condition, which is not the case.

Common Pitfalls:
A common mistake is to attempt guessing different triples without solving the quadratic systematically, which may lead to missing the fact that there is a single solution. Another error is considering negative values of n as valid side lengths. Always remember that triangle side lengths must be positive. Careful algebraic manipulation and factoring prevent these issues.

Final Answer:
There is only one such triple of consecutive positive integers, namely 3, 4 and 5.