Proof problem

[Const.engineer]

Hello, can someone help me with this problem.

a and b are integers, whose remaider is 2 when divided by 3. Proof that there is no x which will solve the equation x^2+x*a+b=0.

 

[Harry_the_cat]

That statement is NOT true.
Here's a counter-example:

Let a=8 and b=11 (Note: when 8 is divided by 3, the remainder is 2, same with 11)
Equation becomes x^2 +8x+11=0
Does this have a solution (real number solution)?
Discriminant is 8^2-4*1*11 >0, so real solutions exist.

Can you give the question exactly as it was given to you?

 

[Const.engineer]

Hello, thank you. So there is no integer x which will solve the equation? Question is exactly the same but x should be an integer which I didn’t mention. Is that enough to proof that there is not integer x to solve the equation in any integers for a and b? (Remainder is 2, when divided by 3).

 

[HallsofIvy]

Hello, can someone help me with this problem.

a and b are integers, whose remaider is 2 when divided by 3.

So a= 3p+ 2 and b= 3q+2 for some integers p and q.

Proof that there is no x which will solve the equation x^2+x*a+b=0.

"Proof" is a noun. The verb is "prove"
\(\displaystyle x^2+ (3p+2)x+ (3q+2)= 0\).
Strictly speaking, the statement here, that there is no x which will solve the equation is false! Every quadratic equation has a solution. If, for example, you take p= q= 1 then a= b= 5. The equation is x^2+ 5x+ 5= 0 which has solutions 5/2+ sqrt(5)/2 and 5/2- sqrt(5)/2. Perhaps they mean "no integer solution". What do you get if you apply the "quadratic formula" or "complete the square?

 

[JeffM]

As has been pointed out, the proposition as you have given it is false. I am going to assume that what is intended is a proof that there is no solution in integers.

Halls of Ivy gave one suggestion to try. Here is another.

[MATH]\exists \text { integers } p \text { and } q \text { such that } a = 3p + 2 \text { and } b = 3q + 2.[/MATH]
Why is that true?

According to the Fundamental Theorem of Algebra

[MATH]\exists \text { numbers } u \text { and } v \text { such that } (x - u)(x - v) = x^2 + ax + b.[/MATH]
[MATH]\therefore u + v = a = 3p + 2 \text { and } uv = 3q + 2.[/MATH]
Now think about a proof by contradiction.