How do we prove that the square of any odd integer is odd?

Question

How do we prove that the square of any odd integer is odd?

Question #59660. Asked by UT-7.

ruteger
Answer has 15 votes

Currently Best Answer

ruteger

Answer has 15 votes.

Currently voted the best answer.

odd number=2k+1 for some integer k

(2k+1)^2=(2k+1)(2k+1)
=4k^2+4k+1 (using FOIL)
=2(2k^2+2k) +1

Therefore the result is also odd, since 2k^2+2k is an integer.

Sep 30 2005, 1:37 AM

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Answer 1

UT-7
Answer has 5 votes

UT-7

Answer has 5 votes.

Perhaps you wanted to say that the result is
odd because it is 1 more than the even integer
4k^2+4k.

Sep 30 2005, 2:13 AM

Answer 2

dim_dude
Answer has 5 votes

dim_dude
19 year member
14 replies avatar

Answer has 5 votes.

Any odd number multipled by another odd number is an odd number. Therefore, if you multiply any odd number with another(itself in this case), then you will obtain an odd square.
I don't get all the formulas ruteger

Sep 30 2005, 7:30 AM

Answer 3

A number is odd if it can be written in the form
2*(an integer)+1,
Eg 7=2*3+1, 13=2*6+1
So in general an odd number can be represented as 2k+1
Squaring this gives 4k^2+4k+1, and you know this is odd because it also can be written in the form 2*(an integer)+1, in this case the integer is 2k^2+2k

Answer 4

UT-7
Answer has 2 votes

UT-7

Answer has 2 votes.

Sorry, dim-dude,but ruteger is right.

Sep 30 2005, 9:17 AM

How do we prove that the square of any odd integer is odd?

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