Let a,b be Natural numbers where a&gt;1. Also p is a ' number. If ax2+bx+c=p for two distinct integer values of x, then prove that ax2+bx+c≠2p for any integral values of x.So this seems to be a pretty straightforward question and I got the final expression that is p=a(x−y)(x−z) where y and z are roots of equations ax2+bx+c−p, so I got p has three factors while a ' number can only have 2 factors. But I couldn’t understand what would happen if x−y=x−z=1. So essentially I need to try and prove that y≠z.

Question

Let a,b be Natural numbers where a&amp;gt;1. Also p is a &#039; number. If ax2+bx+c=p for two distinct integer values of x, then prove that ax2+bx+c≠2p for any integral values of x.So this seems to be a pretty straightforward question and I got the final expression that is p=a(x−y)(x−z) where y and z are roots of equations ax2+bx+c−p, so I got p has three factors while a &#039; number can only have 2 factors. But I couldn’t understand what would happen if x−y=x−z=1. So essentially I need to try and prove that y≠z.

stecchiniror7 · Accepted Answer

Step 1If m, n; m≠n are integer roots to ax2+bx+c=p then ax2+bx+c−p=a(x−m)(x−n)Step 2Now if ax2+bx+c=2p then ax2+bx+c−p=a(x−m)(x−n)=p. If x is an integer and if a&amp;gt;1 and p is &#039; that means a=p and (x−m)(x−n)=1.Step 3And m,n are unequal integer then x−m and x−n are unequal integers. It&#039;s easy to verify that if b,d are integers and bd=1 then b=d=±1.

Let a,b be Natural numbers where \(\displaystyle{a}{>}{1}\).

Answered question

Answer & Explanation

New Questions in Algebra I