prove or disprove the following: the expression n2-n 41 produces a prime number for every positive integer n. hint the following numbers 1, 10, 41.

I think what you probably meant to write is this - please confirm:

n^2 -n +41

It's pretty obvious that for n = 1 you don't get a prime number. QED.

It turns out that for n = 41+a^2, where a = 0, 1, 2, 3,. the value of n^2-n+41 that is not prime but is equal to (41+a(a-1)) x (41+a(a+1))

It was written on the hw package n2-n+41

I need help

What help do you need? I showed that it is not always prime - which disproves the statement that the expression n^2-n+41 produces a prime number for every positive integer.

BTW, you cannot cut and paste math formulas onto this site - otherwise special characters such as "+" get dropped. Also - we use the notation "^" to mean "to the power of." Thus n^2means "n to the second power, or "n squared."

so I would be replace with 10? Because my numbers are 1,10, and 41

(41+10(10-1))x(41+10(10+1))

would this be right?

I think what they wanted was for you try n = 1, n=10 , and n=41. If you do that you get:

For n=1: 1^2 - 1 + 41 = 41, which is a prime number
For n= 10: 10^2 - 10 + 41 = 131, which is a prime number
For n = 41: 41^2 - 41 + 41 = 41^2 = 1681, which is NOT prime

The stuff I added about n= 41+ a^2 is beyond the scope of the question - I just thought it was interesting to point out that there are many values for n that give a non-prime result, starting with n = 41.