Question

Factor
\[
a^{7}+a^{2}+1
\]
and deduce that
1280000401
is composite.

Answer 1

\(\Large \color{MidnightBlue}{\rightarrow a^7 + a^2 + 1^2 }\)

Try to factor it.

Answer 2

did you get that @eliassaab ?

Answer 3

Think about how yo may factor it.

Answer 4

I know how to do it. I posted it just for fun.

Answer 5

eliassaab is a retired maths professor... obviously... this question is for fun only :|

Answer 6

Hint:
Evaluate this quotient
\[\frac{a^7+a^2+1}{a^2+a+1}
\]
using long division.

Answer 7

i did the long division on paper ( awkward on computer)
a^2 + a + 1 is a factor
the quotient is
a^5 - a^4 + a^2 - a + 1

Answer 8

not sure if that will factor

Answer 9

(1+a+a^2) (1-a+a^2-a^4+a^5) <- That's it!~

Answer 10

i'm not sure how to do the second part however i made a guess that 'a' would a number ending in one (pure intuition)

i tried ten but the 10^7 + 10^2 + 1 fell short of the quoted number
tried 20 and it worked out
20^2 + 20 + 1 = 421
this divided into 1280000401 exactly giving 3040381 which = 20^5 - 20^4 + 20^2 - 20 + 1

Answer 11

* correction i guessed that 'a' would end in 0.

Answer 12

I think you got it
\[a^7+a^2+1=\left(a^2+a+1\right)
\left(a^5-a^4+a^2-a+1\right
)
\]

Answer 13

Now, you can answer the second part.

Answer 14

\[20^7+20^2+1=1280000401\]