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

ruteger

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 05, 2:37 AM

UT-7	Perhaps you wanted to say that the result is odd because it is 1 more than the even integer 4k^2+4k. Sep 30 05, 3:13 AM

dim_dude

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 05, 8:30 AM

ruteger

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 Sep 30 05, 8:51 AM

UT-7	Sorry, dim-dude,but ruteger is right. Sep 30 05, 10:17 AM

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