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anonymous
 one year ago
Bet you cant solve this.
anonymous
 one year ago
Bet you cant solve this.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For how many integers n with n <500 can the polynomial \[Pn(x)=x^{6}+n\] be written as a product of two nonconstant polynomials with integer coefficients?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0STOP CHEATING LOL @JoannaBlackwelder

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would 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 nonconstant 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 yaxis. The + n is a positive or negative integer that raises or lowers the parabola along the yaxis 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.
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