Prove that for all integers n, n^2-n+3 is odd. stuck on algebra part :( The question in the book is the following bolded statement. Prove that for all integers n, n^2-n+3 is odd. I rewrote it as this, is that right? For all integers n, there exists an integer such that n^2-n+3 is odd. I'm having problems proving this, the professor said the algebra isn't supose to be hard, but i can't get that. Doing the form of the proof isnt what i'm stuck on though. Heres my start: Proof: Let n be an integer. We must show n^2-n+3 is an odd integer. Since n is an integer, then n(n-1)+3. I don't think i'm doing this the right way, this chapter goes over the quotient remainder theorem. Which is Given any integer n and postie integer d, there exists unique integers q and r such that n = dp + r and 0 =< r < d. From this you can see, that integers can be written in the form, n = 4q or n = 4q + 1 or n = 4q + 2 or n = 4q + 3, for some integer q. Where n = 4q is even, and 4q + 2 is even, and 4q + 1 is odd as well as 4q + 3. This sounds like the right place, becuase the 4q+3 looks really similar to the n(n-1)+3. but i'm not sure what i'm doing becuase I think i can do the following: Since n is an integer, and they say: We must show n^2-n+3 is an odd integer. can i do somthing like this? n^2-n+3 = 4q+3?, where q is an integer. becuase 4q + 3 is an odd integer. Then i come out with: n^2-n = 4q. #but now it seems it lost the odd integer form, and now looks even! n(n-1)=4q; q = n(n-1)/4 But now i'm lost again...any help or suggestions? Thanks!