Contradiction proof about integers

1. (i) Show that \(k ^ { 2 } - 4 k + 5\) is positive for all real values of \(k\).
   (ii) A student was asked to prove by contradiction that "There are no positive integers \(x\) and \(y\) such that \(( 3 x + 2 y ) ( 2 x - 5 y ) = 28\) " The start of the student's proof is shown below.

Assume that positive integers \(x\) and \(y\) exist such that $$\left. \begin{array} { c } ( 3 x + 2 y ) ( 2 x - 5 y ) = 28 \\ \text { If } 3 x + 2 y = 14 \text { and } 2 x - 5 y = 2 \\ 3 x + 2 y = 14 \\ 2 x - 5 y = 2 \end{array} \right\} \Rightarrow x = \frac { 74 } { 19 } , y = \frac { 22 } { 19 } \text { Not integers }$$ Show the calculations and statements needed to complete the proof.

1. Prove by contradiction that there are no positive integers \(p\) and \(q\) such that

$$4 p ^ { 2 } - q ^ { 2 } = 25$$

12 Prove by contradiction that 3 is the only prime number which is 1 less than a square number.

A student was asked to prove by contradiction that "there are no positive integers \(x\) and \(y\) such that \(3x^2 + 2xy - y^2 = 25\)" The start of the student's proof is shown in the box below. Show the calculations and statements that are needed to complete the proof. [4]

A student wishes to prove that, for all positive integers \(a\) and \(b\), \(a^2 - 4b \neq 2\).

1. Prove that \(a^2 - 4b = 2 \Rightarrow a\) is even. [2]
2. Hence or otherwise prove that, for all positive integers \(a\) and \(b\), \(a^2 - 4b \neq 2\). [3]

The three sides of a right-angled triangle have lengths \(a\), \(b\) and \(c\), where \(a, b, c \in \mathbb{Z}\) \includegraphics{figure_6}

1. State an example where \(a\), \(b\) and \(c\) are all even. [1 mark]
2. Prove that it is not possible for all of \(a\), \(b\) and \(c\) to be odd. [3 marks]

Assume that \(a\) and \(b\) are integers such that $$a^2 - 4b - 2 = 0$$

1. Prove that \(a\) is even. [2 marks]
2. Hence, prove that \(2b + 1\) is even and explain why this is a contradiction. [3 marks]
3. Explain what can be deduced about the solutions of the equation $$a^2 - 4b - 2 = 0$$ [1 mark]

Prove by contradiction that there are no positive integers \(a\) and \(b\) with \(a\) odd such that $$a + 2b = \sqrt{8ab}$$ [4]

If \(a\) and \(b\) are odd integers such that 4 is a factor of \((a - b)\), prove by contradiction that 4 cannot be a factor of \((a + b)\). [3]