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Question 1209349: Find the number of bases n \ge 2$ such that 100_n + 1_n is prime.
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
100n+1n = 101n where n is an integer such that 
101n = 101 base n
To convert 101n to base 10 we will use this quadratic polynomial.
1*n^2 + 0*n^1 + 1*n^0
The coefficients 1,0,1 are from 101n
The exponents 2,1,0 count down by 1
The expression
1*n^2 + 0*n^1 + 1*n^0
simplifies to
n^2 + 1
So 101n converts to n^2+1 base 10.
A few examples are
1012 = 2^2 + 1 = 510
1013 = 3^2 + 1 = 1010
1014 = 4^2 + 1 = 1710
1015 = 5^2 + 1 = 2610
1016 = 6^2 + 1 = 3710
1017 = 7^2 + 1 = 5010
1018 = 8^2 + 1 = 6510
1019 = 9^2 + 1 = 8210
10110 = 10^2 + 1 = 10110
You can confirm each claim with a calculator such as this one
Of that list of examples we see the following primes in base ten: 5, 17, 37, 101
Any prime in base 10 converts to a prime back in base n.
For instance, 1014 is prime (since 1710 is prime). We cannot multiply smaller integer values in base 4 to arrive at 1014
So the original problem is equivalent to asking: "What integer values of n will make n^2+1 prime?"
As far as I know, the problem is unsolved in the mathematics community. Many topics about primes are also unsolved.
Refer to this page and this page for further discussion.
The last link mentions "It is conjectured that this sequence is infinite, but this has never been proved."
It's quite possible that there might be a proof out there somewhere that I haven't found; or a proof could come along later in the future. I'll let another tutor chime in.
Answer by ikleyn(52778)  (Show Source):
You can put this solution on YOUR website! .
Find the number of bases n \ge 2$ such that 100_n + 1_n is prime.
~~~~~~~~~~~~~
This problem, inadvertently, offers to solve a world class mathematical problem
from Number theory (of the Fields prize level) here, passing by.
Perhaps, a great thinker is behind this post, who wishes that somebody will solve it for him soon . . .
Bravo ! Plus an elementary proof of Fermat's last theorem, too.
So as not to get up twice.
To do this, we need to revive Euler, Dirichlet, Riemann and five more world class level mathematicians,
to provide them all conditions for work of the Google corporation level for 20 - 40 years,
with quantum super-computers and free food, publishing special journals/magazines
and by holding annual conferences on the subject.
And, perhaps, establish contact with extraterrestrial civilizations in other galaxies.
. . . . . . . . . . . .
In some countries, in collections of preparatory problems for high-level Math Olympiads,
it is customary to print one or two unsolved mathematical problems - simply to inspire young readers.
Naturally, nobody of publishers does not expect that such problems will be solved,
but the tradition is the tradition.
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