Question

HI,how to prove (1/5)n^5+(1/3)n^3+(7/15)n is a integer,when n is any intege.
and i think this question is equal to 15|3n^5+5n^3+7n,but i can't finish it.

Answer 1

yes, you are right to judge that the expression is
(3n^5+5n^3+7n)/15.
Now use the mathematical induction technique, i.e
1. Take n=1
[3(1)^5+5(1)^3+7(1)]/15=15/15=1 (an integer of course!)
2.Let the statement is true for n=k(where k is a positive integer), that is,
(3k^5+5k^3+7k)/15 is an integer.
3. We'll now prove that the statement is true for n=k+1.
That is,
[3(k+1)^5+5(k+1)^3+7(k+1)]/15
=[3(k^5+5k^4+10k^3+10k^2+5k+1)+5(k^3+3k^2+3k+1)+7(k+1)]/15
=(3k^5+15k^4+30k^3+30k^2+15k+3+5k^3+15k^2+15k+5+7k+7)/15
Isolating the first,7th and11th terms from the rest in the numerator we get another equivalent expression:
=(3k^5+5k^3+7k)/15+(15k^4+30k^3+45k^2+30k+15)/15
=(3k^5+5k^3+7k)/15+(k^4+2k^3+3k^2+2k+1). Done. As there is no doubt that the expression inside the 2nd parentheses is an integer.So is the one inside the 1st according to our supposition in 2 above(where we have supposed that this expression is an integer).