How to prove this ?
#2|k^2# then #2|k# for some #k\inZZ#
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"How to prove this ?
#2|k^2# then #2|k# for some #k\inZZ#"
Proof 1: Suppose that #2 | k^2#, but #2 \cancel(|) k#. This means that #k^2# is even, and #k# is odd. But the square of ad odd number is still odd, since
#(2k+1)^2 = 4k^2+4k+1#
so, it is impossible that #k# is odd and #k^2# is even.
Proof 2: If a number is a perfect square, all the exponents in its prime factorization must be even. Since #k^2# is a perfect square, its prime factorization must be like
#k^2 = 2^m * r#
where #m# is even and #r# is the rest of the factorization. In fact, #2# divides #k^2# by hypothesis, and #m# must be even because #k^2# is a square. Since #k=sqrt(k^2)#, the factorization of #k# will contain #2^{m/2}#, which means that #2# divides #k#.
Again, we're shown that it is impossible that #2# divides #k^2# without dividing #k#.
