Discovering Integer Solutions to Equations: Prime or Not?

[mr_coffee]

Hello everyone.

I'm suppose to prove this but I'm having troubles figuring out how u find "distinct" integers. Meaning they can't be the same number. i figured it out they just wanted integers though. Here is the question:
There are distinct integers m and n such that 1/m + 1/n is an integer.

I wrote:
Let m = n = 1. Then m and n are integers such that 1/m + 1/n = 1/1 + 1/1 = 2, which is an integer.

Is there a processes to figuring these things out or is it a guessing game?

Also a harder one is this one:
There is an integer n such that 2n^2-5n+2 is prime.

I looked up what the defintion of a prime number is and i got the following:
An integer n is prime if and only if n > 1 and for all positive integers r and s, if n = (r)(s), then r = 1, or s = 1.

So i wasn't sure where to start with that so I tried to factor 2n^2+5n+2 to see what happens and i got: (x-2)(2x-1). x = 2 or x = 1/2. Because 1/2 is not greater than 1 (x = 1/2) does this mean the whole expression is also not prime? Is that enough to prove it? IT says there IS an integer n that makes that expression prime though.

 

[StatusX]

-1 is an integer. And remember, you're not trying to find a root of 2n^2-5n+2, you're looking for an n such that this evaluates to a prime number. Your factorization will help: try to make one of the factors 1 and the other a prime number.

 

[HallsofIvy]

Do you have trouble with the word "distinct"? 1 and 1 are not distinct!

 

[mr_coffee]

Thanks for the help guys!
By letting m = 1, and n = -1, u get 0 which is an integer. For the 2nd one, am I allowed to just let as you suggested, (n-2)(2n-1); n-2 = 1, 2n-1 = 3;
n = 3, or n = 2. If you plug in 3 for n, u get 2(3)^2-5(3)+2 = 5, which is prime. So by example this is true?