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

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The equation n^2 + n^3 = n^4 - n^2(n - 1) holds true only for n = 2, as confirmed through algebraic manipulation. Initial exploration with different bases like 3 and 4 led to incorrect results, but revealed a pattern in the calculations. The breakdown of the equation shows that after dividing by n^2, simplifying leads to the conclusion that 2n = n^2, which only satisfies n = 2. This indicates that the relationship is unique to the number 2 within the natural numbers. The discussion highlights the importance of verifying mathematical patterns through substitution and simplification.
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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?.
 
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It looks like you are wondering for what values of $n\in\mathbb{N}$ is the following true:

$$n^2+n^3=n^4-n^2(n-1)$$

Divide through by $n^2\ne0$:

$$1+n=n^2-(n-1)$$

Distribute:

$$1+n=n^2-n+1$$

Add $n-1$ to both sides:

$$2n=n^2\implies n=2$$

So, the only value of $n$ for which that works is $n=2$. :)
 
MarkFL said:
It looks like you are wondering for what values of $n\in\mathbb{N}$ is the following true:

$$n^2+n^3=n^4-n^2(n-1)$$

Divide through by $n^2\ne0$:

$$1+n=n^2-(n-1)$$

Distribute:

$$1+n=n^2-n+1$$

Add $n-1$ to both sides:

$$2n=n^2\implies n=2$$

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