Solving a Number Theory Problem Using Fermat's Little Theorem

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To solve the problem of finding 2^70 + 3^70 mod 13, Fermat's Little Theorem can be applied to reduce the exponent from 70 to 10, since 13 is prime. The calculations begin with evaluating 2^2 mod 13, which equals 4, and 3^2 mod 13, which equals 9. The discussion highlights the use of algebraic identities for simplifying expressions, particularly noting that a^n + b^n can be expressed in terms of a^n - (-b)^n when n is odd. This approach aids in further simplifying the problem to find the final result. The conversation emphasizes the importance of understanding these mathematical identities in solving number theory problems.
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Homework Statement


http://math.stanford.edu/~vakil/putnam07/07putnam2.pdf

I am working on number 2.
So I want to find 2^70 + 3^70 mod 13.
I can use Fermat's Little Theorem to reduce the exponent to 10, but I do not know what to do next...


Homework Equations





The Attempt at a Solution

 
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2^2 = 4 and 3^2 = 9

and 4^5 + 9^5 = (4+9)*something.
 
morphism said:
4^5 + 9^5 = (4+9)*something.
Is that true? Where does that come from?
 
You know how there's a formula for a^n - b^n? Well, there's also one for a^n + b^n when n is odd. (a^n + b^n = a^n - (-b)^n.)
 
I see. Thanks.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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