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

  • Context: MHB 
  • Thread starter Thread starter Angel11
  • Start date Start date
Click For Summary
SUMMARY

The equation \( n^2 + n^3 = n^4 - n^2(n-1) \) holds true exclusively for \( n = 2 \). Through algebraic manipulation, dividing by \( n^2 \) and simplifying leads to the conclusion that \( 2n = n^2 \), confirming that \( n \) must equal 2. Attempts to generalize this equation for other integers, such as 3 and 4, do not yield valid results, reinforcing that 2 is the sole solution.

PREREQUISITES
  • Understanding of algebraic manipulation and simplification
  • Familiarity with polynomial equations
  • Knowledge of natural numbers and their properties
  • Basic skills in mathematical proofs
NEXT STEPS
  • Study polynomial identities and their proofs
  • Explore the properties of natural numbers in algebra
  • Learn about mathematical induction techniques
  • Investigate other algebraic equations and their solutions
USEFUL FOR

Students, educators, and mathematicians interested in algebraic identities and problem-solving techniques involving natural numbers.

Angel11
Messages
11
Reaction score
0
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?.
 
Mathematics news on Phys.org
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 :)
 

Similar threads

  • · Replies 59 ·
2
Replies
59
Views
133K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
5
Views
2K