Identifying errors in proofs

5 Charlie claims to have proved the following statement.
"The sum of a square number and a prime number cannot be a square number."

1. Give an example to show that Charlie's statement is not true. Charlie's attempt at a proof is below.
   Assume that the statement is not true.
   ⇒ There exist integers \(n\) and \(m\) and a prime \(p\) such that \(n ^ { 2 } + p = m ^ { 2 }\). \(\Rightarrow p = m ^ { 2 } - n ^ { 2 }\) \(\Rightarrow p = ( m - n ) ( m + n )\) \(\Rightarrow p\) is the product of two integers. \(\Rightarrow p\) is not prime, which is a contradiction.
   ⇒ Charlie's statement is true.
2. Explain the error that Charlie has made.
3. Given that 853 is a prime number, find the square number \(S\) such that \(S + 853\) is also a square number.

7. (i) Given that \(a\) and \(b\) are integers such that $$a + b \text { is odd }$$ Use algebra to prove by contradiction that at least one of \(a\) and \(b\) is odd.
(ii) A student is trying to prove that $$( p + q ) ^ { 2 } < 13 p ^ { 2 } + q ^ { 2 } \quad \text { where } p < 0$$ The student writes: $$\begin{gathered} \qquad \begin{array} { c } p ^ { 2 } + 2 p q + q ^ { 2 } < 13 p ^ { 2 } + q ^ { 2 } \\ 2 p q < 12 p ^ { 2 } \\ \text { so as } p < 0 \quad 2 q < 12 p \\ q < 6 p \end{array} \end{gathered}$$ a. Identify the error made in the proof.
b. Write out the correct solution.

8 Kai is proving that \(n ^ { 3 } - n\) is a multiple of 3 for all positive integer values of \(n\). Kai begins a proof by exhaustion.
Step 1 $$n ^ { 3 } - n = n \left( n ^ { 2 } - 1 \right)$$ Step 2 When \(n = 3 m\), where \(m\) is a \(n ^ { 3 } - n = 3 m \left( 9 m ^ { 2 } - 1 \right)\) non-negative integer which is a multiple of 3 Step 3 When \(n = 3 m + 1\), $$\begin{aligned} & n ^ { 3 } - n = ( 3 m + 1 ) \left( ( 3 m + 1 ) ^ { 2 } - 1 \right) \\ & = ( 3 m + 1 ) \left( 9 m ^ { 2 } \right) \\ & = 3 ( 3 m + 1 ) \left( 3 m ^ { 2 } \right) \end{aligned}$$ Step 5 Therefore \(n ^ { 3 } - n\) is a multiple of 3 for all positive integer values of \(n\) 8

1. Explain the two mistakes that Kai has made after Step 3. Step 4 \section*{which is a multiple of 3
   which is a multiple of 3}
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   别
   all positive integer values of \(n\) \section*{a} \includegraphics[max width=\textwidth, alt={}]{b7df05bf-f3fc-4705-a13c-6b562896fa9f-10_58_49_1037_370} 墐 pount \(\_\_\_\_\) \(\_\_\_\_\) " \(\_\_\_\_\) 8
2. Correct Kai's argument from Step 4 onwards.

5 A student's attempt to prove by contradiction that there is no largest prime number is shown below.
If there is a largest prime, list all the primes.
Multiply all the primes and add 1.
The new number is not divisible by any of the primes in the list and so it must be a new prime. The proof is incorrect and incomplete.
Write a correct version of the proof.

A student was attempting to prove that \(x = \frac{1}{2}\) is the only real root of $$x^3 + \frac{1}{4}x - \frac{1}{2} = 0.$$ The attempted solution was as follows. $$x^3 + \frac{1}{4}x = \frac{1}{2}$$ $$\therefore \quad x(x^2 + \frac{1}{4}) = \frac{1}{2}$$ $$\therefore \quad x = \frac{1}{2}$$ or $$x^2 + \frac{1}{4} = \frac{1}{2}$$ i.e. $$x^2 = -\frac{1}{4} \quad \text{no solution}$$ $$\therefore \quad \text{only real root is } x = \frac{1}{2}$$

1. Explain clearly the error in the above attempt. [2]
2. Give a correct proof that \(x = \frac{1}{2}\) is the only real root of \(x^3 + \frac{1}{4}x - \frac{1}{2} = 0\). [3]

The equation $$x^3 + \beta x - \alpha = 0 \quad \text{(I)}$$ where \(\alpha\), \(\beta\) are real, \(\alpha \neq 0\), has a real root at \(x = \alpha\).

1. Find and simplify an expression for \(\beta\) in terms of \(\alpha\) and prove that \(\alpha\) is the only real root provided \(|\alpha| < 2\). [6]

An examiner chooses a positive number \(\alpha\) so that \(\alpha\) is the only real root of equation (I) but the incorrect method used by the student produces 3 distinct real "roots".

1. Find the range of possible values for \(\alpha\). [7]

\(a\) and \(b\) are two positive irrational numbers. The sum of \(a\) and \(b\) is rational. The product of \(a\) and \(b\) is rational. Caroline is trying to prove \(\frac{1}{a} + \frac{1}{b}\) is rational. Here is her proof: Step 1 \quad \(\frac{1}{a} + \frac{1}{b} = \frac{2}{a + b}\) Step 2 \quad \(2\) is rational and \(a + b\) is non-zero and rational. Step 3 \quad Therefore \(\frac{2}{a + b}\) is rational. Step 4 \quad Hence \(\frac{1}{a} + \frac{1}{b}\) is rational.

1. 1. Identify Caroline's mistake. [1 mark]
   2. Write down a correct version of the proof. [2 marks]
2. Prove by contradiction that the difference of any rational number and any irrational number is irrational. [4 marks]

A student notices that when he adds two consecutive odd numbers together the answer always seems to be the difference between two square numbers. He claims that this will always be true. He attempts to prove his claim as follows: Step 1: Check first few cases \(3 + 5 = 8\) and \(8 = 3^2 - 1^2\) \(5 + 7 = 12\) and \(12 = 4^2 - 2^2\) \(7 + 9 = 16\) and \(16 = 5^2 - 3^2\) Step 2: Use pattern to predict and check a large example \(101 + 103 = 204\) subtract 1 and divide by 2 for the first number Add 1 and divide by two for the second number \(52^2 - 50^2 = 204\) it works! Step 3: Conclusion The first few cases work and there is a pattern, which can be used to predict larger numbers. Therefore, it must be true for all consecutive odd numbers.

1. Explain what is wrong with the student's "proof". [1 mark]
2. Prove that the student's claim is correct. [3 marks]