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Thursday, September 2, 2021

2021/069) 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.

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

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