2
$\begingroup$

Prove that if $t$ is a natural number then there exists a natural number $n > 1$ such that $(n, t) = 1$ and none of the numbers $n + t, n^2 + t, n^3 + t…$ are perfect powers.

There is a solution posted at AOPS, by considering $n = t(t+1)^2 + 1$. Two cases are discussed: $t + 1$ is not a perfect power and $t + 1$ is a perfect power.

I have no problem understand the first case: if $t + 1$ is not a perfect power, then $n^k + t$ can be expressed in the form of $(t + 1)(b(t + 1) + 1)$, since $(t + 1)\nmid(b(t + 1) + 1)$, $(t + 1)$ has to be perfect power if $n^k + t$ is a perfect power. But this is contradiction.

But I got lost when trying to understand the second case, especially from "Consider $n^k+t=c^d$ so by proof...", can anyone help?

$\endgroup$
2
  • $\begingroup$ It means that $t+1$ should be a perfect power of some number $l$ by the same method used in the first case. Since we assumed $m$ not to be a perfect power, $l$ has to be a power of $m$, so $d$ divides $r$. The last line is just straightforward. It is a proof by contradiction. $\endgroup$
    – Henry
    Nov 27, 2013 at 6:17
  • $\begingroup$ That solves 99% of my doubts, but why $(c - n_0)(c^{d-1}+...+n_0^{ks(d-1)})>=n_0$? It must be very simple, I am just so ignorant. $\endgroup$
    – Mathchoice
    Nov 27, 2013 at 6:58

1 Answer 1

Reset to default
4
$\begingroup$

The problem comes from a not so exact translation.

The claim is that that $n$ works, and to prove it, let assume the contrary, that is, there exist $k,c,d$ s.t. $n^k+t=c^d$ then it continues the same.

I hope it makes it clear.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .