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I understand that this is a bi-conditional proof and have been able to successfully solve the backwards direction, but am having difficulty proving the forwards direction. This is my work I have done for my attempt at a direct proof:

For the first part, we assume that $n^2+2n+3$ is even. By the definition of an even integer, there exists some integer $a$ such that $n^2+2n+3 = 2a$.We can manipulate both sides by subtracting the quantity $4n+10$. This will result in the equation $n^2-2n-7 = 2a-4n-10$.

I opted to stop writing this proof because there was no way of getting $4 | (n^2−2n− 7)$. As you can see, currently, I can factor out a 2 and use a variable and the closure properties of integers as well as the definition of divides to get $2 | (n^2−2n− 7)$.

I have also tried to use a proof by contraposition. I used the division algorithm with $4 \nmid (n^2−2n− 7)$ to show that $n^2+2n+3$ is odd, but when I used a remainder of two, I got that the $n^2+2n+3$ was even, which is also wrong.

How would I be able to show the forward direction?

confusedwproofs
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2 Answers2

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Welcome to Mathematics Stack Exchange.

$n^2+2n+3$ is even $\iff n^2+3$ is even $\iff n^2$ is odd $\iff n$ is odd

$\iff n-1$ is even $\iff 4|(n-1)^2=n^2-2n+1\iff 4|n^2-2n+1-8=n^2-2n-7$

J. W. Tanner
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$$n^2-2n-7=(n-1)^2-8$$ will be divisible by $4$

iff $4|(n-1)^2$

iff $2|(n-1)\iff n$ is odd

If $n$ is odd, $$n^2+2n+3=(n+1)^2+2$$ is even

Again $n^2+2n+3$ will be even

iff $n^2$ is odd

iff $n$ is odd which is the necessary and sufficient condition for

$$4|(n^2-2n-7)$$

lab bhattacharjee
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