Challenge For Champions !!! 1

We can write

$a^{k}-1=(a-1)(a^{k-1}+a^{k-2}+ \dotsb +a+1)$ ............(1)

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$\textbf{Case 1}$ - For $k>1$ ,

we have $\quad\quad\quad\quad(a^{k-1}+a^{k-2}+ \dotsb +a+1)>1$ .....................(2)

But $a^{k}-1$ is a prime so only 1 and $a^{k}-1$ are its factors.

From (1) and (2),we get

$a-1=1 \implies a= 2$

Now $2^{k}-1$ is a prime. We can easily prove that $k$ must be prime.

$\quad \quad\quad\quad$ Because $k<15$ , Possible values of 'k' are $k=2/3/5/7/11/13$

but for $k=11$ ; $2^{11}-1=2047=23\times 89$

so for $a=2$ we have $k=2/3/5/7/13$

$\therefore$ Total 5 pairs $(a,k)$ .

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$\textbf{Case 2}$ - For $k=1$

we have $a-1=prime$

$a <98 \implies a-1 <97$

So $\displaystyle 24$ values of $a$ are possible. (There are 24 primes less than 97)

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Total no of pairs $(a,k)$ is $24 +5 =\boxed{29}$