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 [tex]a^2 - 4b \neq 2[/tex]
Let [tex]a^2 - 4b = 2[/tex]
If a is an even integer, then it can be written as [tex]a = 2c[/tex], then,
[tex]a^2 - 4b = 2\\(2c)^2 - 4b =2\\4(c^2 -b) = 2\\(c^2 -b) =\frac{1}{2}[/tex]
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 [tex]a = 2c+1[/tex], then,
[tex]a^2 - 4b = 2\\(2c+1)^2 - 4b =2\\4(c^2+c-b) = 2\\(c^2+c-b) =\frac{1}{4}[/tex]
RHS is a fraction but LHS can never be a fraction, thus it is impossible.
Thus, our assumption was wrong and [tex]a^2 - 4b \neq 2[/tex].
