A question about powers n^2+n^3=n^4−n^2(n−1)

[Angel11]

So recently i was plaing around with powers again and i found that (2^2)+(2^3)=(2^4)-(2^2) which when i thought of that at the start i thought it was an error since i didn't use a calculator but after it i confirmed that it is right. So after that i wanted to see if the same combination of powers works withother numbers instead of 2. for example (3^2)+(3^3)=(3^4)-(3^2). Which of course was wrong although it brought me to another conglusion which was "(2^2)+(2^3)=((2^4)-(2^2)*1)),(3^2)+(3^3)=((3^4)-(3^2)*2), (4^2)+(4^3)=((4^4)-(4^2)*3)... So my question is how does this work?.

 

[MarkFL]

It looks like you are wondering for what values of $n\in\mathbb{N}$ is the following true:

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

Divide through by $n^2\ne0$:

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

Distribute:

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

Add $n-1$ to both sides:

\(\displaystyle 2n=n^2\implies n=2\)

So, the only value of $n$ for which that works is $n=2$. :)

 

[Angel11]

It looks like you are wondering for what values of $n\in\mathbb{N}$ is the following true:

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

Divide through by $n^2\ne0$:

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

Distribute:

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

Add $n-1$ to both sides:

\(\displaystyle 2n=n^2\implies n=2\)

So, the only value of $n$ for which that works is $n=2$. :)

Thank you for explaning :)