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Verify by the method of contradiction p : If a,`b in Z` , then `a^2-4b ne 2`

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Suppose this is false . So , there exists integers , a, `b in Z` for which
`a^2-4b=2 " " ...(i)`
`rArr a^2=4b+2`
`rArr a^2=2(2b+1)`
So, `a^2` is even , also a is even
Let a =2c for some integer c
Putting the value of (a) in (i)
`(2c)^2-4b=2`
So, `4c^2-2b=1`
`rArr 2c^2-2b=1`
`rArr 1=2(c^2-b) " "...(ii)`
(ii) tells us that 1 is even , which is not possible
So, our intial assumption is wrong and p is true
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