if n is a positive integer greater than 2 and p and q are prime numbers greater than 6 then what is common factor of n^2-n and p^2-q^2???
@oobleck
Answer is 7
To find the common factor of n^2 - n and p^2 - q^2, we need to factorize both expressions and see if any factors are common.
Let's start by factorizing n^2 - n:
n^2 - n = n * (n - 1)
Next, let's factorize p^2 - q^2 using the difference of squares formula:
p^2 - q^2 = (p + q) * (p - q)
Now, we can look for the common factors by comparing the factorizations of n^2 - n and p^2 - q^2.
The common factors of n^2 - n and p^2 - q^2 are the factors that appear in both factorizations.
From the factorization of n^2 - n, we have n and (n - 1).
From the factorization of p^2 - q^2, we have (p + q) and (p - q).
So, the common factors are n and (n - 1), as well as (p + q) and (p - q).
Therefore, the common factors of n^2 - n and p^2 - q^2 are n, (n - 1), (p + q), and (p - q).
well, n^2-n = n(n-1) which is even
the p^2-q^2 is odd-odd which is also even
so 2 is clearly a common factor
huh? Suppose n=3. 3^2-3 = 6
and suppose p and q are 11 and 7. 11^2-6^2 = 72
I don' see no steenking 7s in thos numbers.