Unique Prime Numbers

$a^{k} + 1 = b^{m + 1} \Rightarrow a^{k} = b^{m + 1} - 1 \Rightarrow a^{k} = (b - 1)(b^{m} + b^{m - 1} + ... + 1)$ .
$b - 1 < b + 1 \le b^{m} + b^{m - 1} + ... b + 1$ , since $m \ge 1$ . Note that $b - 1$ and $b^{m} + b^{m - 1} + ... b + 1$ are both powers of $a$ , since $a$ is a prime. Therefore, $b - 1 \mid b^{m} + b^{m - 1} + ... b + 1$ .

But $0 \equiv b^{m} + b^{m - 1} + ... b + 1 \equiv (b^{m} - 1) + (b^{m - 1} - 1) + ... + (b - 1) + m + 1 \pmod{b - 1}$ . Since $b^{k} - 1 = (b - 1)(b^{k - 1} + b^{k - 2} + ... b + 1)$ , so $b - 1 \mid b^{k} - 1$ , for $k \in \mathbb{N}$ .

$(b^{m} - 1) + (b^{m - 1} - 1) + ... + (b - 1) + m + 1 \equiv m + 1 \pmod{b - 1}$ , so $\boxed{b - 1 \mid m + 1}$ .

a) If $m + 1$ is a prime,

I) If $m = 1$ ,

$a^{k} = b^{2} - 1 = (b + 1)(b - 1)$ . Since $(b + 1) - (b - 1) = 2$ , and both are powers of $a$ , so $2 = a^{j} - a^{k - j} = a^{k - j}(a^{2j - k} - 1)$ , where $j \le k$ and $b + 1 = a^{j}$ . Since both factors are positive integers, one of them is $1$ and the other is $2$ .

If $a^{k - j} = 1, a^{2j - k} - 1 = 2$ , then $k = j$ , and consequently $a^{k} = 3, a = 3, k = 1, m = 1, b = 2$ . If $a^{k - j} = 2, a^{2j - k} - 1 = 1$ , then $a = 2, k - j = 1$ , so $k = j + 1$ . $2^{j - 1} - 1 = 1 \Rightarrow 2^{j - 1} = 2 \Rightarrow j = 2 \Rightarrow k = 3$ , so $a = 2, k = 3, m = 1, b = 3$ .

So, this gives two solutions for $a$ , namely $a = 2, 3$ .

II) If $m > 1$ ,

From $b - 1 \mid m + 1$ , and that $m + 1$ is a prime, we conclude that $b = 2$ or $b = m + 2$ .

i) $b = 2$ ,

Then $2^{m + 1} - 1 = a^{k}$ .

Suppose $k > 1$ , then $2^{m + 1} - 1$ is a perfect prime power, and $m + 1$ is a prime. But from the fact that the p-th Mersenne number is never a perfect prime power if p is a prime , we have a contradiction.

So, $k = 1$ , and $a = 2^{m + 1} - 1$ , which is a Mersenne prime. We can check that if $a$ is a Mersenne prime less than $1000$ , $m = log_{2}(a + 1) - 1$ , b = 2, k = 1) works. (Note that $n \in \mathbb{N}$ .)

ii) $b = m + 2$

Then $(m + 2)^{m + 1} - 1 = (m + 1)[(m + 2)^{m} + (m + 2)^{m - 1} + ... + 1] = a^{k}$ .

Since $m + 1$ is a prime and a power of $a$ , it follows that $a = m + 1$ ,

$(m + 2)^{m + 1} - 1 = (m + 1)^{k}$

Consider the equation mod $m + 2$ ,

$(m + 2)^{m + 1} - 1 \equiv -1 \pmod{m + 2}$ and $(m + 1)^{k} \equiv (-1)^{k} \pmod{m + 2}$ , so $(-1)^{k} \equiv -1 \pmod{m + 2}$ . Note that $-1$ is not congruent to $1 \pmod{m + 2}$ , so $k$ is odd.

But $(m + 2)^{m + 1} = (m + 1)^{k} + 1 = (m + 2)[(m + 1)^{k - 1} - (m + 1)^{k - 2} + ... - (m + 1) + 1]$ , since k is odd. So $(m + 1)^{k - 1} - (m + 1)^{k - 2} + ... - (m + 1) + 1$ is divisible by $m + 2$ .

But $0 \equiv (m + 1)^{k - 1} - (m + 1)^{k - 2} + ... - (m + 1) + 1 \equiv (-1)^{k - 1} - (-1)^{k - 2} - (-1) + 1 \pmod{m + 2}$ But the last expression is congruent to $k$ mod $m + 2$ , since $k$ is odd, so $m + 2 \mid k$ , which means $k$ is even. (Since $m + 2$ is even.), a contradiction!

b) If $m + 1$ is composite,

Then $m + 1 = pq$ , for some prime $p$ and some positive integer $q$ greater than $1$ . Therefore, $b^{m + 1} = (b^{q})^{p}$ is a perfect p-th power. But we have shown that there are no solutions when $p$ is a prime and $k \ne 1, 3$ .

When $k = 3$ , then the only solution is $a = 2, k = 3, p = 2, b^{q} = 3$ , giving $q = 1$ , a contradiction. When $k = 1$ , then the only solution for $b^{q}$ is $2$ , which again gives $q = 1$ , a contradiction.

Therefore, the only solutions for $a$ are $2$ and the Mersenne primes less than $1000$ , giving $\boxed{2, 3, 7, 31, 127}$ and thus the answer is $2 + 3 + 7 + 31 + 127 = \boxed{170}$ .

If $k > 1$ , Mihăilescu's theorem states that the only possible solution for the equation $a^k + 1 = b^{m+1}$ is $(a,k,b,m) = (2,3,3,1)$ .

If $k = 1$ , then we can rewrite the equation as $a = b^{m+1}-1 = (b-1)(b^m+b^{m-1}+\dots+1)$ . Since $a$ is a prime number, we must have $b-1 = 1$ or $b^m+b^{m-1}+\dots+1 = 1$ . The latter is impossible, so $b$ must be equal to $2$ . Checking whether $a = 2^{m+1}-1$ is prime or not for $m$ from $1$ to $8$ (the largest $m$ for which $2^{m+1} < 1000$ is $8$ ), we get that the possible values for $a$ are $3, 7, 31, 127$ .

So, the sum of all possible values of $a$ is $2 + 3 + 7 + 31 + 127 = 170$ .

We will distinguish two cases.

Case 1. b=2. Then $a$ is odd. Since $m+1\geq 2$ , by looking at the equation $a^k+1=2^{m+1}$ modulo $4$ we conclude that $a\equiv 3 \pmod 4$ and $k$ is odd. Since $k$ is odd, we get $a^k+1=(a+1)(a^{k-1}-...+1)$ . So $a+1$ is a power of $2$ , say $a+1=2^s.$ This implies that $a^k+1=1+ (2^s-1)^k= 1+(-1)+k2^s-\frac{k(k-1)}{2}2^{2s}+...\equiv k2^s \pmod {2^{2s}}$

Because $k$ is odd, this implies that $m+1=s$ , which forces $k$ to be $1$ . So the only possibility in this case is that $a$ is a prime of the form $2^s-1$ . Because $a<1000$ , $s\leq 9$ . Checking powers of $2$ , we get the following values for $a$ : $3,7,31,127$ .

Case 2. b>2. By replacing $b$ with a suitable power, we will assume that $m+1$ is prime. Rewrite the equation as $a^k=(b-1)(b^m+...+b+1)$ .

Note that $1<b-1<b^m+...+1$ , so both $b-1$ and $b^m+...+1$ are divisible by the prime $a$ . Since $b-1\equiv 0 \pmod a,$ we have $b\equiv 1 \pmod a,$ so $b^m+...+1\equiv (m+1) \pmod a.$ As we assumed that $m+1$ is prime, we have $m+1=a.$ So our equation becomes $a^k+1=b^a$ Moreover, $b-1$ is a power of $a$ , so $b=a^s+1$ for some $s\geq 1$ . Therefore $b^a-1=(1+a^s)^a-1=a\cdot a^s + \frac{a(a-1)}{2}\cdot a^{2s}+...$ If $s\geq 2,$ this is congruent to $a^{s+1}$ modulo $a^{2s}$ and is greater than $a^{s+1}$ . So $s=1$ , thus $b=a+1$ . If $a\geq 3,$ then $(a+1)^a-1=a^2+\frac{a(a-1)}{2}a^2+...$ is congruent $a^2(1+\frac{a(a-1)}{2})$ modulo $a^3$ and is greater than that. So it cannot be a power of $a$ . Therefore, $a=2$ , $b=3$ . So we get $2^3+1=3^2$ , which is a solution.

Putting this all together, we get the answer as $2+3+7+31+127=170.$

i found out the solution by substituting values for a,b,k and m.