Powers function n^5+n^4=(n^5−n^3)(n−1)−(n−2)

[Angel11]

Hello again,i am back. So now i have diffrent questions while spending my time in powers for some reason that i can't explain...Anyways let's get to the point. these ones are really simular to the previous one.
n^{5}+n^{4}=(n^{5}-n^{3})*(n-1)-(n-2), n^{4}+n{3}=(n^{4}-n^{2}*(n/(n-1)). You can try to replace n by any number which is >0. And my question is:"WHY does this happen (i mean it goes a little bit into philosophical thinking rather than logical but i would like an explanation like why and 3 digit number that has the same number in every digit if you add it together and multiply the result by 37 it becomes the same number).(Whew) Also if you had the patience to read all of this and acctoully came up with an answer i want to say you are a legend (Yes)

 

[Greg]

Hint: $ 3 \cdot 37 = 111 $

 

[MarkFL]

Let's look at your first equation:

\(\displaystyle n^5+n^4=\left(n^5-n^3\right)(n-1)-(n-2)\)

If we use $n=1$, we get:

\(\displaystyle 2=1\)

So, it's not an identity...and we would have to use a root finding technique to get approximations for the real values of $n$ for which the equation is true:

\(\displaystyle n\approx0.88157969798488504310\)

\(\displaystyle n\approx2.6222866129011998799\)

Let's look at the second equation you gave (I assume it is the following):

\(\displaystyle n^4+n^3=\left(n^4-n^2\right)\left(\frac{n}{n-1}\right)\)

\(\displaystyle n^4+n^3=\frac{n^5-n^3}{n-1}\)

Since you stated $0<n$ we may divide through by $n^3$ to obtain:

\(\displaystyle n+1=\frac{n^2-1}{n-1}\)

Multiply through by $n-1$:

\(\displaystyle (n+1)(n-1)=n^2-1\)

So, given that this is an identity, you original equation is as well, with the additional restriction $n\ne1$. :)

 

[Greg]

Alternatively,

$$ \left( n^4 - n^2 \right) \left( \frac{n}{n - 1} \right), \, n \gt 0, \, n \ne 1 $$

$$ = \frac{ n^5 - n^3 }{ n - 1 } $$

$$ = \frac{ n^3(n^2 - 1) }{ n - 1 } $$

$$ = \frac{ n^3(n + 1)(n - 1) }{ n - 1} $$

$$ = n^3(n + 1) $$

$$ = n^4 + n^3 $$

 

[topsquark]

\(\displaystyle n^{5}+n^{4}=(n^{5}-n^{3})*(n-1)-(n-2)\)

I suspect that it is supposed to be
\(\displaystyle n^5 + n^4 = (n^5 - n^3) \left ( \frac{n - 1}{n - 2} \right )\)

similar to the other one. However now we have no solutions for n, integer or otherwise.

-Dan

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Hint: $ 3 \cdot 37 = 111 $

Okay I'm officially clueless, a state which I have a lot of experience in. How is this a hint?

-Dan

 

[HallsofIvy]

Okay I'm officially clueless, a state which I have a lot of experience in. How is this a hint?

-Dan

The original question was ":"WHY does this happen (i mean it goes a little bit into philosophical thinking rather than logical but i would like an explanation like why and 3 digit number that has the same number in every digit if you add it together and multiply the result by 37 it becomes the same number)."
For example, if we add the digits in 555 we get 15 and 555/37= 15. If we add the digits in 888 we get 24 and 888/37= 24.

That is, "why, if we have something like aaa, where "a" is a single digit, if you add them, you get "3a", while if you divide by 37, you also get "3a"". And greg1313's response was that aaa= a(111)= a(3)(37).