Do there exist three integers in Arithmetic Progression whose product is prime ?
If Yes, then what are the three integers and if No, then why ?
[Note: The numbers: x1, x2, x3, x4, x5, x6,........ are said to be in Arithmetic Progression if (x2 - x1) = (x3 - x2) = (x4 - x3) = (x5 - x4) = ........ and so on].
Let the destined prime number be denoted by p.
If the arithmetic sequence has common difference 0, then it consists of a triple of the same integer, i.e., (x,x,x), say.
Then, x^3 =p gives x=(p)^(1/3), which is NOT an integer. Contradiction.
Accordingly, the three term arithmetic sequence must be given by:
(x,y,z) where, x<y<z.
Since any prime number is >1, it follows that:p>1, so that: -1< -p
Hence by trial and error, we observe that:
x*y*z=p is satisfied only when:
(x,y,z)=(-p,-1,1)
=> z-y=1-(-1)=2
Thus, the arithmetic sequence has a common difference of 2, so that: x=-1-2=-3
Then, x*y*z=(-3)*(-1)*1=3
Consequently, 3 is the only prime number that satisfies all the conditions of the problem, and the requisite three integers are: -3, -1 and 1.
Edited on December 16, 2021, 11:31 pm
Edited on December 16, 2021, 11:32 pm
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