Question 1209349
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100<sub>n</sub>+1<sub>n</sub> = 101<sub>n</sub> where n is an integer such that {{{n >= 2 }}}


101<sub>n</sub> = 101 base n


To convert 101<sub>n</sub> 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 101<sub>n</sub>
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 101<sub>n</sub> converts to n^2+1 base 10.


A few examples are
101<sub>2</sub> = 2^2 + 1 = 5<sub>10</sub>
101<sub>3</sub> = 3^2 + 1 = 10<sub>10</sub>
101<sub>4</sub> = 4^2 + 1 = 17<sub>10</sub>
101<sub>5</sub> = 5^2 + 1 = 26<sub>10</sub>
101<sub>6</sub> = 6^2 + 1 = 37<sub>10</sub>
101<sub>7</sub> = 7^2 + 1 = 50<sub>10</sub>
101<sub>8</sub> = 8^2 + 1 = 65<sub>10</sub>
101<sub>9</sub> = 9^2 + 1 = 82<sub>10</sub>
101<sub>10</sub> = 10^2 + 1 = 101<sub>10</sub>
You can confirm each claim with a calculator such as <a href="https://www.rapidtables.com/convert/number/base-converter.html">this one</a>


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, 101<sub>4</sub> is prime (since 17<sub>10</sub> is prime). We cannot multiply smaller integer values in base 4 to arrive at 101<sub>4</sub>


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 <a href="https://math.stackexchange.com/questions/44126/primes-of-the-form-n21-hard">this page</a> and <a href="https://oeis.org/A002496">this page</a> 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. 
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