Show that if 3n + 5 is odd, then n^2 +1 is even.

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We are asked to show that if 3n+5 is odd, then `n^2+1 ` is even.

If 3n+5 is odd, then we can write 3n+5=2k+1 for some integer k. So 3n=2k-4 or 3n=2(k-2). Thus 3n is even which implies that n is even. For even n, `n^2+1 ` is always odd since the square of an even number is even.

Thus the conjecture as stated is false.

Consider this counterexample: let n=2. Then 3(2)+5=11 which is odd but `2^2+1=5 ` which is also odd.

If 3n+5 is even then n is odd and the square of n plus one is even. We say that 3n+5 and `n^2+1 ` have the same parity: if one is even then the other expression represents an even number also, or both are odd.


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