Let `a ,b in N and a > 1.` Also `p` is a prime number. If `a x^2+b x+c=p` for any intergral values of `x ,` then prove that `ax^2+b x+c!=2p` for any integral value of `xdot`

Explanation

Given `ax^(2) + bx + c - p = 0 ` has integral roots. Let alpha, beta be the roots. Then
`ax^(2) + bx + c - p = a(x - alpha ) ( x - beta)`
Now from `ax^(2) + bx + c = 2p`, we have " "(1)
`ax^(2) + bx + c - p = p`
or ` a(x - alpha) (x - beta)` = p" " [Using (1)]
In the above equation, L.H.S. has three factors but R.H.S. is prime number, which is contradiction . Hence , `ax^(2) + bx + c = 2p` connot have integral roots .