Determine the value of the exponent a that satisfies the equation 9^a + 40^a = 41^a, where all bases are positive integers.

Introduction / Context:This question connects exponents with the geometry of Pythagorean triples. The numbers 9, 40, and 41 suggest a relation similar to a right triangle with sides forming a triple. The task is to find an exponent a such that 9^a + 40^a equals 41^a. Instead of trying random large exponents blindly, we can look for patterns and test small integer exponents.

Given Data / Assumptions:

• The equation is 9^a + 40^a = 41^a.
• All bases 9, 40, and 41 are positive integers.
• a is a real exponent, but we look for integer values among the options.
• Options provided are 1, 2, 3, 4, and 5.

Concept / Approach:Recognise that 9, 40, and 41 resemble a Pythagorean triple because 9^2 + 40^2 = 41^2. This suggests that a = 2 might satisfy the equation exactly. We test a = 2 first. If it works, there is no need to try higher values. In general, for a larger exponent, the largest base dominates, making equality very unlikely unless there is a special relationship like the one for squares.

Step-by-Step Solution:Test a = 2. Compute 9^2 = 81.Compute 40^2 = 1600.Add them: 81 + 1600 = 1681.Compute 41^2: 41^2 = 1681 as well.Thus 9^2 + 40^2 = 41^2, so a = 2 satisfies the equation exactly.There is no need to test other exponents because this provides a perfect equality.

Verification / Alternative check:To check that other exponents do not work, note that for a = 1 we have 9^1 + 40^1 = 49 while 41^1 = 41, so the left side is greater. For a = 3, 9^3 + 40^3 = 729 + 64000 = 64729, while 41^3 = 68921, and the right side is greater. As a increases, the term with the largest base 41 will dominate more strongly, so 41^a will grow faster than 9^a + 40^a. Therefore, equality occurs only at a = 2 among positive integer exponents.

Why Other Options Are Wrong:

• a = 1 gives 49 on the left and 41 on the right, not equal.
• a = 3, 4, and 5 produce values where 41^a is larger than the sum of the other two terms.
• None of these satisfy the required equality.
• Only a = 2 produces an exact match between left and right sides.

Common Pitfalls:

• Not recognising the Pythagorean triple pattern and trying random exponents without a plan.
• Making calculation errors while squaring or cubing the numbers.
• Assuming that if one exponent works, others might also work, without checking the growth behaviour of exponential functions.

Final Answer:2