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Assume that $a$ and $b$ are integers such that a^2 - 4b - 2 = 0 9 (a) Prove that $a$ is even - AQA - A-Level Maths Pure - Question 9 - 2022 - Paper 3

Question 9

Assume that $a$ and $b$ are integers such that a^2 - 4b - 2 = 0 9 (a) Prove that $a$ is even. 9 (b) Hence, prove that $2b + 1$ is even and explain why this is a c... show full transcript

Worked Solution & Example Answer:Assume that $a$ and $b$ are integers such that a^2 - 4b - 2 = 0 9 (a) Prove that $a$ is even - AQA - A-Level Maths Pure - Question 9 - 2022 - Paper 3

Step 1

9 (a) Prove that $a$ is even.

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Answer

To prove that $a$ is even, start with the given equation:

$a^2 - 4b - 2 = 0$

Rearranging gives:

$a^2 = 4b + 2$

Note that $4b$ is always even since it is a multiple of $4$ , and $2$ is also even. Therefore, the right-hand side ( $4b + 2$ ) must be even.

Since the left-hand side ( $a^2$ ) equals an even number, we deduce that $a^2$ must be even. This implies that $a$ must also be even because the square of an odd integer is odd.

Step 2

9 (b) Hence, prove that $2b + 1$ is even and explain why this is a contradiction.

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Answer

From the previous step, we established that $a$ is even. Now substitute $a$ as follows:

$a = 2k$

for some integer $k$ . Substituting into our equation yields:

$(2k)^2 - 4b - 2 = 0$

Calculating gives:

$4k^2 - 4b - 2 = 0$

Rearranging leads to:

$4k^2 - 2 = 4b$

Dividing through by $2$ results in:

$2k^2 - 1 = 2b$

This implies that:

$2b + 1 = 2k^2$

Since $2k^2$ is even, $2b + 1$ must also be odd. This means that $2b + 1$ being odd contradicts the original assumption that both $b$ and $a$ are integers, leading to a conclusion that the assumption of $a$ as even must be incorrect.

Step 3

9 (c) Explain what can be deduced about the solutions of the equation $a^2 - 4b - 2 = 0$.

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Answer

Considering the deductions made in the previous parts, particularly part (b), we conclude that there are no integer solutions to the equation

a^2 - 4b - 2 = 0.

This is because it leads to a contradiction regarding the integer nature of $2b + 1$ . Thus, the only valid conclusion is that no integers $a$ and $b$ can satisfy this equation.

Assume that $a$ and $b$ are integers such that a^2 - 4b - 2 = 0 9 (a) Prove that $a$ is even - AQA - A-Level Maths Pure - Question 9 - 2022 - Paper 3

Worked Solution & Example Answer:Assume that $a$ and $b$ are integers such that a^2 - 4b - 2 = 0 9 (a) Prove that $a$ is even - AQA - A-Level Maths Pure - Question 9 - 2022 - Paper 3

9 (a) Prove that $a$ is even.

9 (b) Hence, prove that $2b + 1$ is even and explain why this is a contradiction.

9 (c) Explain what can be deduced about the solutions of the equation $a^2 - 4b - 2 = 0$.

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