For a given positive integer $n>1$ , how to find all positive integers $s,t$ such that $n^s-(n-1)^t=1$ ? $s=t=1$ is clearly a solution . One more thing is clear that for any such $s,t$ we must have $1+(n-1)^t=n^s=(1+(n-1))^s \ge 1+(n-1)^s$ , so $t \ge s$ . Now if $t=s >1$ , then $n^s-(n-1)^s >1$ , so if $t>s>1$ , then $n^s=1+(n-1)^t\ge1+(n-1)^{s+1} \ge 1+(n/2)^{s+1} >(n/2)^{s+1}$ so $2^s >n$ i.e. $s > \log_2 n$ . But I cannot get anything else ; Please help , thanks in advance .
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You said "$2^s >n$", but it's $2^{s+1}>n$. – user26486 Apr 01 '15 at 15:06
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Similar problems: $2^n\pm1=x^k$, $3^k-1=x^n$ and $2^n - 1 = a^k$. – Bart Michels Apr 02 '15 at 09:04
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If n=3, s=2 and t=3 then we have a solution.
Perhaps this is the only one?
user227093
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Yes , I know about Catalan-Mihăilescu's theorem ; but I think it is too a heavy a tool to attack the present problem – Apr 01 '15 at 15:00
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@SaunDev But it's the exact same problem, heavy or not. Isn't it? – peter.petrov Apr 01 '15 at 15:03
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@peter.petrov No, Catalan's conjecture concerns the more general $n^s-m^t=1$ instead of $n^s-(n-1)^t=1$. – Bart Michels Apr 01 '15 at 17:19