I'm trying to prove by MI. I have already distributed n+1, but now I'm stuck on how I can show 9 divides the RHS since $42n$ and $3n^3$ does not divide evenly.
$$(n+1)^3+(n+2)^3+(n+3)^3=3n^3+18n^2+42n+36$$
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HINT:
Induction:
Let $f(n)=n^3+(n+1)^3+(n+2)^3$
$f(m+1)-f(m)=(m+3)^3-m^3=9(m^2+3m+3)$
So, $f(m+1)$ will be divisible by $9\iff f(m)$ is.
Check for the base case
lab bhattacharjee
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Is there a rule that if 9|p(x) then 9|p(x+1)- p(x) – RustyShackleford Oct 01 '14 at 13:48
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@wannabe_n00b, No, if any number $n$ divides $a-b$ and $b,n$ must divide $a$ – lab bhattacharjee Oct 01 '14 at 16:32
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$$(n-1)^3+n^3+(n+1)^3=3n^3+6n=3(n^3-n)+9n$$
Now $n^3-n=(n-1)n(n+1)$ is product of three consecutive integers, hence divisible by $3!$
Also, $$(n+1)^3+(n+2)^3+(n+3)^3=3n^3+18n^2+42n+36\equiv3(n^3-n)\pmod9$$
lab bhattacharjee
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Could I prove this by simply showing mod 9 on p(n) and p(n+1) is the same, therefore p(n+1) is divisible by 9 – RustyShackleford Oct 01 '14 at 14:20
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@wannabe_n00b, Exactly. That's what is http://mathworld.wolfram.com/Congruence.html – lab bhattacharjee Oct 01 '14 at 16:33
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I didn't know you could have a negative remainder with modulus operator – RustyShackleford Oct 01 '14 at 17:32
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@wannabe_n00b, Conventionally for $r=n%m 0\le r<m$. But, it is sometimes useful to set $-\dfrac m2<r\le \dfrac m2$ – lab bhattacharjee Oct 01 '14 at 18:05
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your term is equivalent to
$3(n+1)(n^2+2n+3)$
if $n\equiv 0\mod 3$ then $n^2+2n+3\equiv 0 \mod 3$
if $n\equiv 1 \mod 3$, then $n^2+2n\equiv 0 \mod 3$
if $n\equiv 2 \mod 3$ then $n+1 \equiv 0 \mod 3$
Dr. Sonnhard Graubner
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If the property is true for $n$, then it is true for $n+9$ (all three cubes will increase by a multiple of 9). So it suffices to check for $n=0,1,...8$.
$$0^3+1^3+2^3=9=1\cdot 9$$ $$1^3+2^3+3^3=36=4\cdot 9$$ $$2^3+3^3+4^3=99=11\cdot 9$$ $$3^3+4^3+5^3=216=24\cdot 9$$ $$4^3+5^3+6^3=405=45\cdot 9$$ $$5^3+6^3+7^3=684=76\cdot 9$$ $$6^3+7^3+8^3=1071=119\cdot 9$$ $$7^3+8^3+9^3=1584=176\cdot 9$$ $$8^3+9^3+10^3=2241=249\cdot 9$$
Actually, as you showed that the sum is a multiple of $3$, it even suffices to show that the average is a multiple of $3$ for $n=0, 1, 2$. $$\frac{0^3+1^3+2^3}3=3=1\cdot 3$$ $$\frac{1^3+2^3+3^3}3=12=4\cdot 3$$ $$\frac{2^3+3^3+4^3}3=33=11\cdot 3$$
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So if I factor out the 3, it will be proof that it is divisible by 9? – RustyShackleford Oct 01 '14 at 14:00
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If you factor out the $3$ you need to prove that the remaining factor is a multiple of $3$: $9|3x\iff 3|x$. In your case, just prove that $n^3+14n$ is a multiple of $3$ for $n=0, 1, 2$. – Oct 01 '14 at 14:01
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$$9\Big|\Big(n^3 + (n+1)^3 + (n+2)^3\Big)\iff9\Big|\Big(3n^3+9n^2+15n+9\Big)\iff3\Big|(n^3+5n)\iff3\Big|\Big(n(n^2+5)\big).$$ It is obviously true when $3|n$.
If the property is true for $n$, then it is true for $n+3$, so it only remains to check that for $n=1,2$, $3\Big|(n^2+5)$.
$$3|6,\\3|9.$$