Call the middle number a, and call the common difference d. Both a and d must be integers.
Your numbers are a-d, a, and a+d.
Their product must be (a-d)(a)(a+d)
This is (a^2 - d^2)(a).
But let's think about it. (a^2 - d^2)(a) is a prime...and yet it has 'a', an integer, as a factor! That means that either:
1) a^2 - d^2 = 1, and a is a prime.
2) a=1, and a^2 - d^2 is a prime.
3) a^2 - d^2 = -1, and a is a negative prime number.
4) a=-1, and a^2 - d^2 is a negative prime number.
The first case gives a^2 = 1 + d^2. The only integer solutions to that are a=+/-1, d=0. That gives the semi-trivial sequence of 1, 1, 1, and that product is not a prime.
The second case means that 1 - d^2 must be prime. But the smallest prime is 2, and 1 - d^2 can't equal 2 since d^2 is positive.
The third case gives d^2 - a^2 = 1. This gives d=+/-1, and a=0 for integer solutions. This contradicts the statement that 'a' is a negative prime number.
Cross your fingers!
The last case gives a=-1, and a^2 - d^2 is the negative of some prime. Thus d^2 - a^2 must be prime. Plug in, and you get d^2 - 1 being prime. Since d^2 - 1 can be factored into (d+1)(d-1), one of those must be equal to 1. If d+1=1, we get d=0. Bad. If d-1 = 1, then we get d=2.
So we have a=-1 and d=2. Does this work?
Our middle number is -1. That means the first number is -3, and the last number is 1. The product is 3!
Tricky but cool question!
Of course, after all this case-work, I find a quicker way to the solution:
(a-d)(a)(a+d) are three factors. We can't have d=0, because then the sequence would be constant and the product would be a^3. a^3 is never prime, not for a=-1, 0 or 1, and clearly not for bigger numbers.
Thus we have three UNIQUE factors. Only ONE of them is allowed to be a prime number. Therefore, the other two MUST be -1 and 1. Since the product of these three numbers is prime, the third number must be the opposite of a prime. Since -1 and 1 are in the sequence, you just have to count down 2 units to get -3 as the first number.
Your numbers are a-d, a, and a+d.
Their product must be (a-d)(a)(a+d)
This is (a^2 - d^2)(a).
But let's think about it. (a^2 - d^2)(a) is a prime...and yet it has 'a', an integer, as a factor! That means that either:
1) a^2 - d^2 = 1, and a is a prime.
2) a=1, and a^2 - d^2 is a prime.
3) a^2 - d^2 = -1, and a is a negative prime number.
4) a=-1, and a^2 - d^2 is a negative prime number.
The first case gives a^2 = 1 + d^2. The only integer solutions to that are a=+/-1, d=0. That gives the semi-trivial sequence of 1, 1, 1, and that product is not a prime.
The second case means that 1 - d^2 must be prime. But the smallest prime is 2, and 1 - d^2 can't equal 2 since d^2 is positive.
The third case gives d^2 - a^2 = 1. This gives d=+/-1, and a=0 for integer solutions. This contradicts the statement that 'a' is a negative prime number.
Cross your fingers!
The last case gives a=-1, and a^2 - d^2 is the negative of some prime. Thus d^2 - a^2 must be prime. Plug in, and you get d^2 - 1 being prime. Since d^2 - 1 can be factored into (d+1)(d-1), one of those must be equal to 1. If d+1=1, we get d=0. Bad. If d-1 = 1, then we get d=2.
So we have a=-1 and d=2. Does this work?
Our middle number is -1. That means the first number is -3, and the last number is 1. The product is 3!
Tricky but cool question!
Of course, after all this case-work, I find a quicker way to the solution:
(a-d)(a)(a+d) are three factors. We can't have d=0, because then the sequence would be constant and the product would be a^3. a^3 is never prime, not for a=-1, 0 or 1, and clearly not for bigger numbers.
Thus we have three UNIQUE factors. Only ONE of them is allowed to be a prime number. Therefore, the other two MUST be -1 and 1. Since the product of these three numbers is prime, the third number must be the opposite of a prime. Since -1 and 1 are in the sequence, you just have to count down 2 units to get -3 as the first number.
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-3, -1, 1
Product is 3.
Edit: AH, very nice analysis. I have to admit I just did it by inspection.
Product is 3.
Edit: AH, very nice analysis. I have to admit I just did it by inspection.