Problems & Puzzles: Puzzles

Puzzle 900. Why 25? Sebastián Martín Ruiz sent the following puzzle. Q. Why the numerator of the reduced quotient concerning to the product expression {Π[((pi)^2+1)/(pi-1)]} from i=1 to n, always end in "25" for n>2? n=3, 325/2 n=4, 8125/6 n=5, 99125/6 n=6, 8425625/36 n=7, 1221715625/288 ...

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Contribution came from Emmanuel Vantieghem

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Emmanuel wrote:

Writing the nth prime as p(n) and Ruiz's product as

   
x(n) = Product{for i = 1 to n}  of (p(i)^2 + 1)/(p(i) - 1) 

then I verified the conjecture for all n < 500000. But the conjecture is certainly true for n big enough.
 

Indeed, p(i)^2 + 1  is divisble by  5  when  p(i)  ends with the digit  3  or  7.

Thus, by the prime number theorem and Dirichlet's theorem on primes in arithmetic progressions, the product{for i = 1 to n}  of (p(i)^2 + 1)  will be divisible by 5^(n/2).

On the other hand, p(i)-1  is divisible by  5  only when  p(i)  ends with the digit  1.

Thus, by the two theorems mentioned, the product{for i = 1 to n}  of  p(i) - 1  will at least be divisible by 5^(n/4)

Thus, on the average, the numerator of  x(n)  will be divisible by  5^(n/4).

But, it is easy to see that the numerator of  x(n)  is allways odd :

the fraction  (p(i)^2 + 1)/(p(i) - 1)  has numerator even, but not divisible by 4 while the denominator is even and eventually divisible by 4.

Included, you may find two graphics in this pdf file.

It is not possible that the numerator of  x(n)  ends in 75 since that numerator must be of the form  4m+1. It is indeed the product of sums of two squares and hence also a sum of two squares, whence all its odd prime divisors must be of the form  4m+1. (a number that ends with 75 is always of the from 4m+3)

With this addition, the Ruiz’s observation is still a conjecture that could be false? (CR)

As far as I know, it is indeed a conjecture. But the chance that it fails is extremely small. This is suggested in the graphs that I sent you. It is also possible that my knowledge about the distribution of primes is not big enough.

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