Evaluations for input

So I need to find the least odd prime factor of ( 2019^8 + 1 ). Hmm, that seems pretty big, but I think I can break it down step by step.

First, I know that finding prime factors of large exponents can be tricky, but there might be some properties or theorems that can help simplify the process. Since the expression is ( 2019^8 + 1 ), maybe I can look for patterns or ways to factor this expression.

I recall that expressions of the form ( a^n + b^n ) can sometimes be factored, especially when ( n ) is a power of 2. In this case, ( n = 8 ), which is ( 2^3 ), so that might be useful.

Let me see if I can factor ( 2019^8 + 1 ). I know that ( a^4 + 1 ) can be factored as ( (a^4 + 1) = (a^2 + \sqrt{2}a + 1)(a^2 - \sqrt{2}a + 1) ), but that involves irrational numbers, which isn't helpful for factoring into integers. Maybe there's another way.

Wait, perhaps I can write ( 2019^8 + 1 ) as ( (2019^4)^2 + 1^2 ), which resembles the sum of squares. I know that over the integers, sums of squares don't factor nicely, but maybe there's a different approach.

Another thought: maybe I can consider smaller exponents first and see if there's a pattern or a recurring factor that appears.

Let's try computing ( 2019^2 + 1 ), ( 2019^4 + 1 ), and so on, and see if I can find any common factors or patterns.

First, calculate ( 2019^2 + 1 ):

( 2019^2 = 2019 \times 2019 ). I don't need the exact value right now; maybe I can just keep it as ( 2019^2 ) for now.

So, ( 2019^2 + 1 = 2019^2 + 1 ). Doesn't seem immediately helpful.

Next, ( 2019^4 + 1 ). Again, it's similar to the previous expression, and I don't see an obvious factorization.

Maybe I should try to see if there are any small prime numbers that divide ( 2019^8 + 1 ). Since I'm looking for the least odd prime factor, I can start testing small primes like 3, 5, 7, 11, and so on, to see if any of them divide ( 2019^8 + 1 ).

Let's start with 3. To check if 3 divides ( 2019^8 + 1 ), I can compute ( 2019^8 + 1 ) modulo 3.

First, find ( 2019 ) modulo 3. Since ( 2019 = 3 \times 673 ), ( 2019 \equiv 0 \pmod{3} ). Therefore, ( 2019^8 \equiv 0^8 \equiv 0 \pmod{3} ), and ( 2019^8 + 1 \equiv 0 + 1 \equiv 1 \pmod{3} ). So, 3 does not divide ( 2019^8 + 1 ).

Next, check 5. Compute ( 2019 ) modulo 5. ( 2019 \div 5 = 403 ) with a remainder of 4, so ( 2019 \equiv 4 \pmod{5} ). Then, ( 2019^8 \equiv 4^8 \pmod{5} ). Since ( 4 \equiv -1 \pmod{5} ), ( 4^8 \equiv (-1)^8 \equiv 1 \pmod{5} ). Therefore, ( 2019^8 + 1 \equiv 1 + 1 \equiv 2 \pmod{5} ), so 5 does not divide ( 2019^8 + 1 ).

Now, check 7. Find ( 2019 ) modulo 7. ( 2019 \div 7 = 288 ) with a remainder of 3, so ( 2019 \equiv 3 \pmod{7} ). Then, ( 2019^8 \equiv 3^8 \pmod{7} ). Let's compute ( 3^8 ) modulo 7.

First, find the pattern of powers of 3 modulo 7:

( 3^1 \equiv 3 \pmod{7} )

( 3^2 \equiv 9 \equiv 2 \pmod{7} )

( 3^3 \equiv 6 \pmod{7} )

( 3^4 \equiv 18 \equiv 4 \pmod{7} )

( 3^5 \equiv 12 \equiv 5 \pmod{7} )

( 3^6 \equiv 15 \equiv 1 \pmod{7} )

Since ( 3^6 \equiv 1 \pmod{7} ), then ( 3^8 = 3^{6+2} = 3^6 \times 3^2 \equiv 1 \times 2 \equiv 2 \pmod{7} ). Therefore, ( 2019^8 + 1 \equiv 2 + 1 \equiv 3 \pmod{7} ), so 7 does not divide ( 2019^8 + 1 ).

Next, check 11. Compute ( 2019 ) modulo 11. ( 2019 \div 11 = 183 ) with a remainder of 6, so ( 2019 \equiv 6 \pmod{11} ). Then, ( 2019^8 \equiv 6^8 \pmod{11} ). Let's find a pattern for powers of 6 modulo 11:

( 6^1 \equiv 6 \pmod{11} )

( 6^2 \equiv 36 \equiv 3 \pmod{11} )

( 6^3 \equiv 18 \equiv 7 \pmod{11} )

( 6^4 \equiv 42 \equiv 9 \pmod{11} )

( 6^5 \equiv 54 \equiv 10 \pmod{11} )

( 6^6 \equiv 60 \equiv 5 \pmod{11} )

( 6^7 \equiv 30 \equiv 8 \pmod{11} )

( 6^8 \equiv 48 \equiv 4 \pmod{11} )

Therefore, ( 2019^8 + 1 \equiv 4 + 1 \equiv 5 \pmod{11} ), so 11 does not divide ( 2019^8 + 1 ).

Continuing this process seems tedious. Maybe there's a better way to approach this problem.

Let me think differently. Perhaps I can consider the expression ( 2019^8 + 1 ) and see if it can be expressed in a form that reveals its prime factors.

I recall that for expressions like ( a^4 + 4b^4 ), there's a Sophie Germain identity: ( a^4 + 4b^4 = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2) ). Maybe I can manipulate ( 2019^8 + 1 ) into a similar form.

Wait, ( 2019^8 + 1 = (2019^4)^2 + 1^2 ). This resembles the sum of squares, but over integers, sums of squares don't factor nicely. Maybe I can consider complex numbers or some other algebraic identities, but that might not help in finding integer prime factors.

Alternatively, perhaps I can consider modulo properties to find possible prime factors. For example, if ( p ) is an odd prime that divides ( 2019^8 + 1 ), then ( 2019^8 \equiv -1 \pmod{p} ), which implies ( 2019^{16} \equiv 1 \pmod{p} ). This means that the order of 2019 modulo ( p ) divides 16, and since ( 2019^8 \equiv -1 \pmod{p} ), the order can't divide 8. Therefore, the order must be 16.

By Fermat's Little Theorem, the order of 2019 modulo ( p ) divides ( p - 1 ). Therefore, ( p - 1 ) must be divisible by 16. So, ( p \equiv 1 \pmod{16} ).

This narrows down the possible primes ( p ) that could divide ( 2019^8 + 1 ) to those primes congruent to 1 modulo 16.

Let me list some primes that are 1 modulo 16: 17, 97, 113, 193, 241, 257, etc.

Now, I can test these primes to see if any of them divide ( 2019^8 + 1 ).

Start with 17. Compute ( 2019 ) modulo 17.

First, find ( 2019 \div 17 ). 17 times 118 is 2006, so ( 2019 - 2006 = 13 ). Therefore, ( 2019 \equiv 13 \pmod{17} ).

Then, ( 2019^8 \equiv 13^8 \pmod{17} ).

Compute ( 13^2 \equiv 169 \equiv 16 \pmod{17} )

( 13^4 = (13^2)^2 \equiv 16^2 \equiv 256 \equiv 1 \pmod{17} )

( 13^8 = (13^4)^2 \equiv 1^2 \equiv 1 \pmod{17} )

Therefore, ( 2019^8 + 1 \equiv 1 + 1 \equiv 2 \pmod{17} ), so 17 does not divide ( 2019^8 + 1 ).

Next, check 97. Compute ( 2019 ) modulo 97.

First, find how many times 97 fits into 2019. 97 times 20 is 1940, and 2019 - 1940 = 79. So, ( 2019 \equiv 79 \pmod{97} ).

Then, ( 2019^8 \equiv 79^8 \pmod{97} ).

This seems a bit cumbersome to compute directly. Maybe I can find a pattern or use Fermat's Little Theorem, which states that ( a^{p-1} \equiv 1 \pmod{p} ) if ( p ) is prime and ( p ) does not divide ( a ).

Here, ( p = 97 ), so ( 79^{96} \equiv 1 \pmod{97} ). But I need ( 79^8 ), which is not directly helpful. Maybe I can compute ( 79^2 ), then ( 79^4 ), and so on.

First, compute ( 79^2 \pmod{97} ):

( 79^2 = 6241 ). Now, divide 6241 by 97.

97 times 64 is 6208, so 6241 - 6208 = 33. Therefore, ( 79^2 \equiv 33 \pmod{97} ).

Next, ( 79^4 = (79^2)^2 \equiv 33^2 \pmod{97} ).

( 33^2 = 1089 ). 97 times 11 is 1067, so 1089 - 1067 = 22. Therefore, ( 79^4 \equiv 22 \pmod{97} ).

Then, ( 79^8 = (79^4)^2 \equiv 22^2 \pmod{97} ).

( 22^2 = 484 ). 97 times 4 is 388, so 484 - 388 = 96. Therefore, ( 79^8 \equiv 96 \pmod{97} ).

Thus, ( 2019^8 + 1 \equiv 96 + 1 \equiv 97 \equiv 0 \pmod{97} ).

So, 97 divides ( 2019^8 + 1 ). Since 97 is the smallest prime congruent to 1 modulo 16 that we've checked and it divides ( 2019^8 + 1 ), it must be the least odd prime factor of ( 2019^8 + 1 ).

Therefore, the least odd prime factor is 97.

Final Answer

[ \boxed{97} ]