SOLUTION: What is the counter example of "The product of a number (n-1) and number (n+1) is always equal to n^2-1?"

Algebra.Com
Question 207728: What is the counter example of "The product of a number (n-1) and number (n+1) is always equal to n^2-1?"
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
If you FOIL out , you will get:


So is true for ALL values of 'n'


This means that "The product of a number (n-1) and number (n+1) is always equal to n^2-1" is ALWAYS true. So once again, we cannot find any counter examples because there are none in this case.


RELATED QUESTIONS

The expression 1/1.2 + 1/1.3 + 1/1.4 +......+ 1/n(n+1) for any natural number n is (a)... (answered by solver91311)
Find a counter-example to the statement 4^n+1 is divisible by 5. Thank... (answered by fractalier)
Consider g(x) = x^n + 4x^n - 1; what is the maximum number of turns it can have? @ The... (answered by edjones)
Dear math teacher, I am having difficulties with the following problem: 4 times nC2 (answered by Theo)
Find the number of bases n \ge 2$ such that 100_n + 1_n is... (answered by math_tutor2020,ikleyn)
2. A student thinks that P(n) = n^3 - n is always a multiple of 6 for all natural numbers (answered by venugopalramana)
Let p(n) and s(n) denote the product and the sum, respectively, of the digits of the... (answered by Edwin McCravy)
P(X < n) is always equal to 1 – P(X > n), regardless of the type of probability... (answered by richard1234)
A number n is said to be perfect it the sum of all its divisors (excluding n itself) is... (answered by ramkikk66)