n^3/3 + n^5/5 + 7n/15 is an integer

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Question from John, a student:

I am very lost on this one:

Prove: For all n in Natural Numbers ( n > 1 ),
n^3/3 + n^5/5 + 7n/15 is an integer

i. Let n=1, then 1/3 + 1/5 + 7/15
= (5 + 3 + 7 )/15 = 15/15 = 1 . True

ii. assume n^3/3 + n^5/5 + 7n/15 is an integer

show this is also true for ( n+1).

I can't figure out how to set this up so that I can use the induction hypothesis to
make the assumption useful in my proof.

Thank You,

Hi John.

Begin with (n+1)³/3 + (n+1)⁵/5 + 7(n+1)/15

= (1/3)( n³ + 3n² + 3n + 1) + (1/5)(n^⁵ + 5n⁴ + 10n³ + 10n² + 5n + 1) + 7n/15 + 7/15
= [ n³/3 + n^⁵/5 + 7n/15 ] + (1/3)( 3n² + 3n + 1) + (1/5)( 5n⁴ + 10n³ + 10n² + 5n + 1) + 7/15
= [an integer] + (1/3)( 3n² + 3n + 1) + (1/5)( 5n⁴ + 10n³ + 10n² + 5n + 1) + 7/15

and you have to demonstrate that the rest is an integer. Can you finish the algebra?

Cheers,
Stephen La Rocque.

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