Question

Easy yet interesting problem !
Show that \(n^{15}+1\) is composite for all integers \(n\ge 2\)

Answer 1

found this while attempting this problem
https://projecteuler.net/problem=421

Answer 2

can be proved easy using cauchy inequality and fundamental theorem of algebra

Answer 3

interesting xD i have a different approach, please show the proof :)

Answer 4

let \(n^5=x\)
\(\large\tt \begin{align} \color{black}{
n^{15}+1\hspace{.33em}\\~\\
=(n^{5})^3+1\hspace{.33em}\\~\\
=(x)^3+1\hspace{.33em}\\~\\
=(x+1)(x^2-x+1)\hspace{.33em}\\~\\
=(n^5+1)(n^{10}-n^5+1)\hspace{.33em}\\~\\
}\end{align}\)

Answer 5

I hope that can be useful

Answer 6

How do you tackle that question though? I don't even know how to begin.

Answer 7

so does that mean n^3+1 is composite ?

Answer 8

*

Answer 9

we can prove a much stronger statement :
\(n^{2k+1}+1\) is composite for all integers \(n\ge 2\) and \(k\ge 1\)

Answer 10

\[n^{2k+1}+1 = n^{2k+1} - (-1)^{2k+1} = (n+1)(n^{2k} - n^{2k-1}+\cdots+1)\]
for \(n\ge 2\), both the factors above are \(\gt 1\) ending the proof.

Answer 11

used this identity for factoring
http://gyazo.com/9a998e02beaead2b99413acd559c3267

Answer 12

this derived from binomial theorem

Answer 13

so this gives
sum of squares composite ,sum of cubes composite ,....ect

Answer 14

i still think ,then why we can write primes ad difference of squares :P

Answer 15

sum of squares need not be composite

Answer 16

4^2 + 1^2 is not composite

Answer 17

i see
but it is when the second term is not 1

Answer 18

how do you prove that ?

Answer 19

http://en.wikipedia.org/wiki/Proofs_of_Fermat's_theorem_on_sums_of_two_squares

Answer 20

eh , ok i guess i was wrong

Answer 21

\(n = a^2 + b^2\) is a prime only when \( n\equiv 1 \pmod 4 \)

Answer 22

i see

Answer 23

so this proof :-
n^2=0,1 mod 4
a^2+b^2=0+0=0 mod 4(invalid since it gives a^2+b^2 even)
=0+1=1 mod 4
=1+1=2 mod 4 (invalid since it gives a^2+b^2 even)