Question

Bet you cant solve this.

Answer 1

For how many integers n with |n| <500 can the polynomial \[Pn(x)=x^{6}+n\] be written as a product of two non-constant polynomials with integer coefficients?

Answer 2

your right

Answer 3

https://www.wyzant.com/resources/answers/59611/for_how_many_integers_n_with_ini_500_can_the_polynomial_pn_x_x6_n_be_written_as_a_product_of_two_non_constant_polynomials_with_integer_coefficients

Answer 4

STOP CHEATING LOL @JoannaBlackwelder

Answer 5

its 0 ofc.

Answer 6

would attack this problem along these lines which you can refine.

If the polynomial Pn(x) = (x6 + n) could be written as a product of 2 non-constant polynomials with integer coefficients, then by the Fundamental Theorem of Algebra one of those factors would represent a real root when Pn(x) = 0.

Let's plot Pn(x). First x6 is basically a parabola going thru (x,y) = (0,0) that opens upward and is symmetric about the y-axis. The + n is a positive or negative integer that raises or lowers the parabola along the y-axis leaving x alone.

Hence, for n>=+1, Pn(x) cannot have roots since it always has positive y, and so cannot be expresses in the product format.

For n=0, you get the degenerate case of x6 = 0 with 6 identical x=0 roots. so you can represent x6 = x1x5, x2x4 etc.

For n negative integers, the parabola gets lowered so there are exactly 2 distinct real integer roots:

If n = -1, the roots are x = +- 1 and a factorization is possible
If n = -64, the roots are x = +- 2 and a factorization is possible

If x = +-3, then n would be -729 = - (3)6, which exceeds your |n| < 500 constraint.