Sum of Solutions for a Prime

We see that $b^2-1=(b-1)(b+1)=a^3$ Since $a$ is a prime number, either $b-1=1$ and $b+1=a^3$ , or $b-1=a$ and $b+1=a^2$ .

If $b-1=1$ and $b+1=a^3$ , then $a^3=3$ , which is not possible.

If $b-1=a$ and $b+1=a^2$ , then $a+2=a^2$ , so $a=2,-1$ . Only $a=2$ works, giving $b=3$ , so the answer is $\boxed{3}$ .

Since $a$ is a prime number, the prime factorization of $a^3$ is simply $a^3$ , which has factors $1,a,a^2,a^3$ . Furthermore since $b-1<b+1$ , and both are positive, it follows that the only cases we need to consider are those where $b-1$ is $1$ or $a$ .

Case 1: $b-1=1$ This is invalid since $a^2=3$ which goes not yield an integer solution of $a$ , let alone prime.

Case 2: $b-1=a$ Then $b = a + 1, b+1 = a+2$ and so $a+2 = a^2$ . (Note: We can divide by $a$ as it is non-zero.) This quadratic equation has roots $a = 2, -1$ , and $a=2$ is the only positive prime solution.

The desired sum is therefore 3.

I will write the equation as (b-1)(b+1)=a^3 case 1: a is odd prime number then b must be even to get b-1 and b+1 odd as a^3 is odd so we can't get any solution for this case because we can't get two different odd numbers(b-1) and (b+1) and multiply them to be equal a number power three/ case 2: a is even prime number which is of course 2 by substitution we can get that b=3. Then the sum of all positive numbers will be 3

First consider a=2, which is trivial since it is the only even prime. Substituting gives a=2 and b=3, which is one solution to the problem. We prove now that there are no further solutions. A quadratic residue are the values of q such that x^2 = q mod n Consider quadratic residues modulo 4. When x is even, it can be expressed in the form x = 2k. Therefore x^2 = 4k^2 so it is 0 mod 4. Following a similar argument, when x is odd, it is 1 mod 4. Now consider cubic residues modulo 4. Since all primes beyond 2 are all odd, we can assume that p = 2k + 1. Reduciung (2k+1)^3 modulo 4 gives 2k+2 (mod 4). Now, when k is odd it can be expressed in the form (2m + 1) so 2(2m + 1) + 2 = 4m + 4 which is 0 (mod 4). When k is even, following a similar argument we get 2 (mod 4). So the cubic residues (mod 4) when x is odd are 0 and 2.

We have shown that the quadratic residues (mod 4) are 0 (when x is even) and 1 (when x is odd) and the cubic residues (mod 4) are 0 or 2 (when x is odd). Therefore, since x represents a which is a prime, and it is odd beyond 2, there are no more solutions apart from a = 2 and hence b = 3.

$a^3 +1 = (a + 1) (a^2 - a + 1)$ We're interested only in choices of $a$ which result in perfect squares. So $a^2 - a + 1 = a + 1$ $a^2 - a = a$ $a^2 = 2a$

From this, you can see that this only works if $a = 2$ so $b = 3$

hello,see b^2=a^3+1=(a+1)*(a^2-a+1) and for a is a prime number there is no number to divide a ,so with (a-1) and (a^2-a+1) are also equal which gives only one solution of a=2 and then b=3.

the faster formula for answer it quickly

b^a = a^b + 1

b^2 = a^3 + 1

3^2 = 2^3 + 1 9 = 9

so, the answer is 3

Select least prime and check the value of b if it holds then it is a solution , if it not , then try another prime number for a and check value of b .The process ends when b holds... This problem is solved by trial and error method..

2^3 +1 = 3^2

here you go with some prime numbers:2,3,5,7,11,13,17,19,...... and now check for those primes one by one ....you will find that for only a=2, b can have the value of 3....and when you go forward you would certainly get no other prime for satisfying this equation .....so the only answer is b=3......here see now b take value (-3) also but you say b is a positive number...so the value (-3) can not be the answer....so b=3 .............please give me some more points........thanxxxx....

Start testing numbers. Notice instantly that if solutions are (a, b), then (2, 3) works. As you increase your "b", you start to notice that your "a"s get less and less close to cubed numbers, much less cubed primes. Decide not to continue anymore and try 3 as your final solution just to make sure you're not wasting your time. Rejoice.

b^2=(a^3+1), b=sqrt(a^3+1), now a=2, (2 2 2+1)=9, b=sqrt(9), now b=3

$b^2 = a^3 +1$

$b^2 - 1 = a^3$

$\left(b+1\right)\left(b-1\right) = a^3$

$a$ is prime and so $b+1 = a^2, b-1 = a$

$a^2 - a - 2 = 0$

$\left(a-2\right)\left(a+1\right) = 0$

$a,b > 0$ so $a = 2 \Rightarrow b = 3$

There is only one number for this statement is true

a = 2 ( prime )

$2^{3}$ + 1 = 9

$b^{2}$ = $3^{2}$

b^2 = a^3 + 1 $\rightarrow$ b^2= (a+1)(a^2 -a +1)

this is only has solutions when either (a+1) = (a^2 -a + 1) or when (a+1) is a factor of (a^2-a+1) and (a^2-a+1)/ (a+10) is also a square number

because (a+1) isn't a factor of (a^2-a+1), we only need to consider the first case

(a+1) = (a^2 -a +1) $\rightarrow$ a^2 = 2a where the only solution is a = 2

when a=2, b^3 = 9 and b =3; this is the only positive integer that satisfies the equation

There is only one number that can fit the equation:

$3^2=2^3 + 1$

The answer is 3

If we move the $1$ over, we get $b^2-1 = a^3.$ Notice that we can factor: $(b-1)(b+1) = a^3.$ This is a very important equation, since we know that $a$ is prime. We have two factors multiplying to $a^3,$ and $a^3$ only has one prime factor, $a$ itself. Therefore, if neither of the factors equal $1$ (see below for this case), then both must be divisible by $a.$ (See note $1.$ ) If both are divisible by $a,$ then $a$ must divide their difference. (See note $2.$ ) But their difference is just $2,$ so the only possibility is that $a = 2.$ Substituting this back into the given equation, we find that $b = \sqrt{2^3+1} = 3.$

To finish, we should consider what happens if one of the factors equals $1.$ Setting each factor equal to $1$ separately gives the two values $b = 0, 2.$ The first is invalid because $b$ is positive. We can check the second value in the original equation, to find that it is also invalid. Thus, the only possible $b$ is $b = \boxed{3}.$

Note 1: Specifically, they must be $a$ and $a^2$ in some order, but we didn't need to know that.

Note 2: Say that $b-1 = ma$ and $b+1 = na$ for some positive integers $m, n.$ Then, the difference of these two is $(m-n)a,$ which is a multiple of $a.$ $\square$

perfect cubes are in the form -1, 0, or 1 (mod 9)

Perfect squares are in the form 0, or 1 (mod 4)

Thus the perfect cube has to be congruent to -1(mod 9) so when 1 is added it is 1(mod 4) and an odd number. a^3 +1 will always be even for all primes except for 2.

Thus a=2 is the only solution and b is 3.