Prove No Integers Solve $ax^3+bx^2+cx+d=1$ for x=19,2 for x=62

In summary, the conversation discusses the task of proving that there are no integers a, b, c, and d that can satisfy the equation ax^3+bx^2+cx+d = 1 at x=19 and 2 at x=62. It is shown that this leads to a linear Diophantine equation with the coefficients 231469a+ 3461b+ 53c= 1, which does not have integer solutions. Therefore, it is concluded that it is impossible to find integers a, b, c, and d that satisfy the given conditions.
  • #1
anemone
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Prove that there are no integers $a,\,b,\,c$ and $d$ such that the polynomial $ax^3+bx^2+cx+d$ equals 1 at $x=19$ and 2 at $x=62$.
 
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  • #2
Well, to start with, since $ax^3+ bx^2+ cx+ d$ is 1 when x 19, $a(19)^3+ b(19)^2+ 19x+ d= 6859a+ 361b+ 19c+ d= 1$, And since it is 2 when x= 62, $a(62)^3+ b(62)^2+ a(62)+ d= 238328a+ 3844b+ 62c+ d= 2$,

Subtracting the first from the second, 231469a+ 3461b+ 53c= 1. Since a, b, and c are integers that is a linear Diophantine equation.
 
  • #3
we have $f(62) - f(19) = a (62^3-19^3) + b(62^2 - 19^2) + c(62-19) = 1$
or $(62-19)(a(62^2 + 62 * 19 + 19^2) + b(62+ 19) +c) = 1$
LHS is a multiple of 43 and RHS is 1 so this does not have integer solution
 
  • #4
Country Boy said:
Well, to start with, since $ax^3+ bx^2+ cx+ d$ is 1 when x 19, $a(19)^3+ b(19)^2+ 19x+ d= 6859a+ 361b+ 19c+ d= 1$, And since it is 2 when x= 62, $a(62)^3+ b(62)^2+ a(62)+ d= 238328a+ 3844b+ 62c+ d= 2$,

Subtracting the first from the second, 231469a+ 3461b+ 53c= 1. Since a, b, and c are integers that is a linear Diophantine equation.
because this is a challenge question you are required to answer it fully . this is not a question for help
 

FAQ: Prove No Integers Solve $ax^3+bx^2+cx+d=1$ for x=19,2 for x=62

1. Can this equation be solved for any integer value of x?

No, this equation cannot be solved for any integer value of x. It only has solutions for specific values of x, such as 19 and 62.

2. What is the significance of the number 1 in this equation?

The number 1 is the constant term in this equation and represents the desired outcome or solution.

3. How many solutions does this equation have?

This equation has two solutions, one for x=19 and one for x=62.

4. How does the degree of the polynomial affect the number of solutions?

The degree of the polynomial, in this case being 3, indicates the highest power of x in the equation. For a polynomial of degree n, there can be at most n solutions.

5. Can this equation be solved using methods other than substitution?

Yes, this equation can also be solved using factoring, the rational root theorem, or other algebraic methods. However, substitution is the most straightforward approach for solving this particular equation.

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