Power with odd base and polynomial having no integral or rational solutions The initial task is to prove, using induction, that if $b$ is odd and $n\ge1$, then $b^n$ is odd. I think I got that part. Then, using this fact, you have to show that the polynomial $x^{19}+x+1=0$ has no integral or rational solutions. I don't see the connection. Any hints?
Thank you.
 A: Given the equation
$x^{19} + x + 1 = 0, \tag{0}$
let
$x = \dfrac{p}{q} \tag{1}$
with
$gcd(p, q) = 1; \tag{2}$
then $p/q$ is in lowest terms; i.e., $p$ and $q$ can have no common integral factors other than $\pm 1$.  Substituting (1) into (0):
$(\dfrac{p}{q})^{19} + \dfrac{p}{q} + 1 = 0, \tag{3}$
whence
$p^{19} + pq^{18} + q^{19} = 0; \tag{4}$
this last obtained by multiplying (3) through by $q^{19}$, the denominator of $x$.  Then
$q^{19} = -p^{19} - pq^{18}. \tag{5}$
Now $p$ and $q$ cannot both be even, by (2); if $q$ is odd, then by the assertion which has already been given as proved, $q^{19}$ is odd, and then if $p$ is also odd, $-p^{19} - pq^{18}$ is even, for $p^{19}$ is odd, $q^{18}$ is odd, and $pq^{18}$ is odd, whence the right hand side of (5) is even, a contradiction; clearly if $p$ is even, (5) fails since the right hand side is even.  If $q$ is even, then again by (2) $p$ must be odd, whence $p^{19}$ is odd and $pq^{18}$ is even; then the right hand side of is odd but the left hand side even, another contradiction.  We have exhausted the cases for rational $x = p/q$; there is no rational, hence no integer, solution.  QED.
Note Added Sunday 14 December 2014 11:33 AM PST:  Just to flesh things out, here is a simple inductive proof of the assertion that $b^n$ is odd for odd $b$ and $n \ge 1$:  clearly $b^1 = b$ is odd; assuming $b^k$ is odd, then $b^{k + 1} = bb^k$ is odd since it is the product of two odds.
Also, this little gem should generalize nicely.  How, exactly?  End of Note.
Hope this helps.  Cheers!
And as ever,
Fiat Lux!!!
