Question

Prove that if a and b are integers, then a^2-4b egal or non-egal 2

Answer 1

Answer:

tex]a^2 - 4b \neq 2[/tex]

Step-by-step explanation:

We are given that a and b are integers, then we need to show that $a^2 - 4b \neq 2$

Let  $a^2 - 4b = 2$

If a is an even integer, then it can be written as $a = 2c$, then,

$a^2 - 4b = 2\\(2c)^2 - 4b =2\\4(c^2 -b) = 2\\(c^2 -b) =\frac{1}{2}$

RHS is a fraction but LHS can never be a fraction, thus it is impossible.

If a is an odd integer, then it can be written as $a = 2c+1$, then,

$a^2 - 4b = 2\\(2c+1)^2 - 4b =2\\4(c^2+c-b) = 2\\(c^2+c-b) =\frac{1}{4}$

RHS is a fraction but LHS can never be a fraction, thus it is impossible.

Thus, our assumption was wrong and $a^2 - 4b \neq 2$.