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Prove by induction that for every integer n, $\frac{11n^3 + 25n}{6}$ is an integer (i cant post the actual given expression since it might cause academic offence). I have only worked with proofs with natural numbers so induction was quite linear. For integers how do i go about showing that it is an integer. I tried cases where n = 1, 2, 3, -1, -2, -3 and it is all true but i dont know how to start it. Any insight would be appreciated. Thank you for your time.

Bill Dubuque
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UnKnoWnZ
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    can you show that it's true for $-1$ and if it's true for $-n$ then it's true for $-(n+1)$? – J. W. Tanner May 10 '20 at 04:29
  • Unfortunately, the details of a proof will depend intimately on the concrete expression. We can give some general pointers, but it will mostly be things you have probably heard before. – Arthur May 10 '20 at 04:30
  • It may be easiest to phrase the induction based not on the actual value of the integer, but rather on the absolute value of the integer. Show that if $|x|=k$ then $f(x)$ is an integer. Then use this to show that if $|x|=k+1$ that $f(x)$ is an integer. So $f(0)$ being an integer simultaneously implies $f(1)$ and $f(-1)$ are both integers. These being integers implies $f(2)$ and $f(-2)$ are both integers, etc... Again though, more specific advice depends on the actual expression. – JMoravitz May 10 '20 at 04:31
  • Addition, subtraction and multiplication of integers are always integers. For division you need to prove divisibility. You can use: $P(0) \wedge [\forall n>0 (P(n)\implies P(n+1) \wedge P(-n)\implies P(-n-1))]$ – David P May 10 '20 at 04:33
  • @Arthur I added the expression – UnKnoWnZ May 10 '20 at 04:46
  • @JMoravitz i added the expression – UnKnoWnZ May 10 '20 at 04:48
  • I recommend this answer for any induction question – Ross Millikan May 10 '20 at 04:49
  • For that particular expression you can write $$\frac{11(n+1)^3 + 25(n+1)}{6}=\frac{11n^3 + 25n}{6}+6+11\frac{n(n+1)}{2}.$$ – John B May 10 '20 at 05:05
  • Induction can "go down" just as easily as "go up". If $P(k) \implies P(k\pm 1)$ and $P(c)$ is true for any $c \in \mathbb Z$ then $P(n)$ is true for all $n \in \mathbb Z$. (Because induction goes up, it is true for all $n \ge c$ and because induction goes down it is true for all $n \le c$). – fleablood May 10 '20 at 05:46
  • Since it equals $(n-n^3)/6 + 4n+2n^3$ it suffices to prove $(n-n^3)/6$ is an integer and this is a FAQ, e.g. see the linked dupes. – Bill Dubuque May 10 '20 at 06:05
  • No induction is needed. It suffices to prove that $11n^3 + 25n \equiv 0 \bmod 6$, that is, $11n^3 + 25n \equiv 0 \bmod 2$ and $11n^3 + 25n \equiv 0 \bmod 3$. Both cases are very easy to prove directly. – J.-E. Pin May 10 '20 at 06:11

3 Answers3

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For your particular expression, $$f(n)=\frac{11n^3+25n}{6}.$$

We have the property that $f(-n)=-f(n)$, hence if $f(n)$ is an integer, $f(-n)=-f(n)$ must also be an integer.

Hence, you can focus on proving that for natural number $n$, $f(n)$ is an integer.

For base case, do it for $P(0)$.

Induction step: show that $P(n) \implies P(n+1)$

If you insist: show that $P(n) \implies P(n-1)$ as well. For this particular problem, once you find a relationship between $f(n+1)$ and $f(n)$, both directions should be similar.

I would actually recommend understanding the expression using modulo $6$ arithmetic besides induction.

Siong Thye Goh
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Rewrite:

$$A=11n^3+25n=12n(n^2)-n(n-1)(n-2)$$

$A_1=(n-1) n (n+1)=6 k$

Because it is the product of three consecutive numbers.So A is always divisible by 6,so $A/6$ is integer. To show that by induction put $n+1 $ you get:

$A_1=n(n+1)(n+2)$

which is again a product of three consecutive numbers and is a multiple of 6, that is $A/6$ is integer for both n and n+1 so it is integer for every n.

sirous
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Start anywhere.

If $n = 59$ then $\frac {11n^3 +25n}6 = \frac {11*59^3 + 25*59}6 = \frac {2260644}6 =376774$.

Now prove by induction that we can "go up". If $\frac {11n^2 + 25n}6$ is an integer then prove $\frac {11(n+1)^2+ 25*(n+1)}6$ is an integer.

So $\frac {11(n+1)^3 + 25(n+1)}6 = \frac {11(n^3 + 3n^2 + 3n + 1) + 25n + 25}6=$

$\frac {11n^3 + 25n + (33n^2 + 33n + 36)}6= \frac {11n^3+25n}6 + \frac {11n^2 + 11n + 12}2=$

$\frac {11n^3 +15n}6 + 11\frac {n(n+1)}2 + 6$.

Now $\frac {11n^3 + 15n}6$ is an integer. Ad as either $n$ or $n+1$ is even $11\frac {n(n+1)}2$ is an integer and $6$ is an integer. SO the sum is an integer.

So you have proven if $n =59$ or $n$ is an integer that can be reached from $59$ by adding $1$ any number of times then $\frac {11n^3 + 25n}6$ is an integer.

Now prove by induction that we can "do down. If $\frac {11n^2 + 25n}6$ is an integer then prove $\frac {11(n-1)^2+ 25*(n-1)}6$ is an integer.

So $\frac {11(n-1)^3 + 25(n-1)}6 = \frac {11(n^3 - 3n^2 + 3n - 1) + 25n - 25}6=$

$\frac {11n^3 + 25n + (-33n^2 + 33n - 36)}6= \frac {11n^3+25n}6 + \frac {-11n^2 + 11n - 12}2=$

$\frac {11n^3 +15n}6 -11\frac {n(n-1)}2 - 6$.

Now $\frac {11n^3 + 25n}6$ is an integer. And as either $n$ or $n-1$ is even $11\frac {n(n-1)}2$ is an integer and $6$ is an integer. SO the sum is an integer.

Wo we have proven if $n=59$, or if $n$ can be reached from $59$ by repeatedly adding $1$, or if $n$ can be reached from $59$ by repeatedly subtracting $1$, we have proven our result.

And that covers every integer.

....

We could simplify it.

Start with Base case $n=0$, have the induction step go "both ways" and prove for if true for $n$ then is true for $n\pm 1$.

That's do it.

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or note: $\frac {n^3 + 25n}6$ is an integer if and only if $\frac {(-n)^3 + 25(-n)}6 = -\frac {n^3 + 25n}6$ is an integer.

So we don't have to prove for all $n$. Just all natural $n$.

fleablood
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