Questions: Conjectures and Testing

Question 1

A student computes 1² = 1, 11² = 121, 111² = 12321, and 1111² = 1234321. What conjecture might the student form?

• A. Squaring any number gives a palindrome
• B. The square of a repunit (number with all 1s) produces a palindrome that counts up and then back down
• C. All palindromes are perfect squares
• D. Squaring always produces a number with more digits

Answer: (B)

The pattern shows 1, 121, 12321, 1234321 — each is a palindrome that counts up to the number of digits in the repunit, then counts back down. This is a specific, testable conjecture about repunit squares. Option A is too broad (2² = 4 is not a palindrome). Option C reverses the relationship. Option D is true but not the interesting pattern here.

Question 2

Testing 100 cases that most support a conjecture proves the conjecture is true.

• T. True
• F. False

Answer: False

No number of confirming cases constitutes a proof. The conjecture 'n² + n + 41 is prime for all positive integers n' holds for n = 1 through n = 39, which is impressive — but it fails at n = 40 (40² + 40 + 41 = 41² = 1681, which is not prime in the relevant sense) and obviously at n = 41. Testing builds confidence and is essential for discovering counterexamples, but only a deductive proof establishes truth for all cases.

Question 3

Describe the conjecture-test-refine cycle using an example of a conjecture that needs refinement.

Model answer: Start with the conjecture 'the sum of two prime numbers is always even.' Test: 2 + 3 = 5, which is odd — counterexample found. Refine: 'the sum of two odd prime numbers is always even.' Test: 3 + 5 = 8 (even), 7 + 11 = 18 (even), 13 + 19 = 32 (even). No counterexample found. The refined conjecture is ready for a proof attempt.

The cycle shows how counterexamples drive progress. The original conjecture was too broad — it forgot about 2, the only even prime. The counterexample pointed directly at the problem (one prime was even), which led to the fix (restrict to odd primes). The refined conjecture can then be proven deductively: odd + odd = even.