Solve in ℤ₅ the equation x⁷ + x³ – x +1= 0.

If x ∈ ℤ₅ such that x⁷+x³-x+1=0 then x(x⁶+x²-1)=-1 i.e. x(2x²-1)=4. Now (when I write x=m i mean the class of m modulo 5)

x=0 —> x(2x²-1)=0≠4
x=1 —> x(2x²-1)=1≠4
x=2 —> x(2x²-1)=14=4
x=3 —> x(2x²-1)=1≠4
x=4 —> x(2x²-1)=124=4

so the solutions are the class of 2 and the class of 4 modulo 5.