Say whether the following is true or false and support your answer by a proof: For any integer n, the number n2+n+1 is odd?

3 Answers
Aug 25, 2017

#n^2+n+1# is always odd

Explanation:

If #n# is odd:
#n^2# is also odd
#n^2+n# (sum of two odds) is even
#n^2+n+1# is odd

If #n# is even:
#n^2# is also even
#n^2+n# (sum of two evens) is even
#n^2+n+1# is odd

Aug 25, 2017

#n^2+n+1# is odd

Explanation:

  • Even numbers are numbers of the form #2k# for some integer #k#.

  • Odd numbers are numbers of the form #2k+1# for some integer #k#.

Every integer is either odd or even.

#color(white)()#
Case #n# even

If #n# is even, then there is some integer #k# such that #n = 2k#

Then:

#n^2+n+1 = (2k)^2+2k+1#

#color(white)(n^2+n+1) = 4k^2+2k+1#

#color(white)(n^2+n+1) = 2(2k^2+k)+1#

#color(white)(n^2+n+1) = 2k_1+1#

where #k_1 = 2k^2+k# is an integer.

So #n^2+n+1# is odd.

#color(white)()#
Case #n# odd

If #n# is odd, then there is some integer #k# such that #n = 2k+1#

Then:

#n^2+n+1 = (2k+1)^2+(2k+1)+1#

#color(white)(n^2+n+1) = 4k^2+2k+1+2k+1+1#

#color(white)(n^2+n+1) = 2(2k^2+2k+1)+1#

#color(white)(n^2+n+1) = 2k_2+1#

where #k_2 = 2k^2+2k+1# is an integer.

So #n^2+n+1# is odd.

#color(white)()#
Conclusion

So regardless of whether #n# is even or odd, #n^2+n+1# is odd.

Aug 25, 2017

Is odd.

Explanation:

#n^2+n+1 = n(n+1)+1#

Note that #n(n+1)# is always even so #n(n+1)+1# is odd