30 Divide N5 - N.docx

Show that n5−nn5−n is divisible by 30;30; ∀n∈N∀n∈N

I tried to solve this three-way. And all stopped at some point. I) By induction: testing for 00, 11 and 22 It is clearly true. As a hypothesis, we have 30|n5−n⇒n5−n=30k30|n5−n⇒n5−n=30k. Therefore, the thesis would 30|(n+1)5−n−1⇒(n+1)5−n−1=30j30|(n+1)5−n−1⇒(n+1)5−n−1=30j. By theorem binomial (n+1)5−n−1=n5+5n4+10n3+10n2+5n+1−n−1(n+1)5−n−1=n5+5n4+10n3 +10n2+5n+1−n−1 =30k+5n4+10n3+10n2+5n.=30k+5n4+10n3+10n2+5n. It was a little messy and not given to proceed. II)I tried by Fermat's Little Theorem:

30|n5−n⇒5⋅3⋅2|n5−n30|n5−n⇒5⋅3⋅2|n5−n Analyzing each case, we have; clearly 5|n5−n5|n5−n (Fermat's Little Theorem). Now cases 3|n5−n3|n5−n and 2|n5−n2|n5−n I could not develop.

III)In this I wanted your help especially like to solve using this because this exercise on this handout (Properties of the Greatest Common Divisor). Would show that

gcd(n5−n,30)=30gcd(n5−n,30)=30 Are detailers, please, because'm coursing Theory of the numbers the first time. In the third semester of the degree course in mathematics What would be the idea to solve this by using the Greatest Common? This problem in this handout GCD?

You can easily prove it by induction

(n+1)5−(n+1)=(n5−n)+5n4+10n3+10n2+5n=(n5−n)+5n(n3+2n2+2n+1)(n+1)5− (n+1)=(n5−n)+5n4+10n3+10n2+5n=(n5−n)+5n(n3+2n2+2n+1)

=(n5−n)+5n(n+1)(n2+n+1)=(n5−n)+5n(n+1)(n2+n+1) =(n5−n)+5n(n+1)((n−1)2−3n)=(n5−n)+5n(n+1)((n−1)2−3n) =(n5−n)+5n(n+1)(n−1)2−15n2(n+1)=(n5−n)+5n(n+1)(n−1)2−15n2(n+1)  

is divisible by 3030 by the inductive assumption n(n+1)(n−1)n(n+1)(n−1) is the product of three consecutive integers, thus divisible by 22 and 33.

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n(n+1)n(n+1) is even.

Added If you use induction probably the best way to approach the problem is by observing that the inductive step reduces to

30|5n4+10n3+10n2+5n=5(n4+2n3+2n2+n)30|5n4+10n3+10n2+5n=5(n4+2n3+2n2 +n) which is equivalent to

6|n4+2n3+2n2+n6|n4+2n3+2n2+n Treat this as a new induction problem.

Note that the degree of the polynomial decreased by 11, and the inductive step, if not obvious, is also a similar problem but the degree again one less. Repeat until you get an obvious statement...