Are any of the following statements is correct :
i) 2 | a ⇒ 2| a^2
ii) 2 | a^2 ⇒ 2 | a
iii) 2 | a ⇔ 2 | a^2
for integers a ?

Question

Are any of the following statements is correct :
i) 2 | a ⇒ 2| a^2
ii) 2 | a^2 ⇒ 2 | a
iii) 2 | a ⇔ 2 | a^2
for integers a ?

anonymous · Answer

They look all true.

anonymous · Answer

If you think a statement is false, try to produce a counterexample.

anonymous · Answer

@jayzdd but how do I show it?

anonymous · Answer

like in the first one$$a = 2*x \rightarrow a*a=2*x$$
and because 2 is a divider to a it must be a divider to a*a right?

nincompoop · Answer

HA?

anonymous · Answer

@nincompoop that equal sign is if I should prime factorize a, and we know that 2 is a divider of a?

anonymous · Answer

@nincompoop could you explain instead?

anonymous · Answer

@welshfella

amistre64 · Answer

(2n+1)(2n+1) = 4n^2+4n+1 = 2(2n^2+2n) + 1 = 2k+1

an odd squared, is odd

but then since 2 doesnt divide odd numbers, it doesnt divide either factor to start with

even*odd, divides the even number, and therefore the product and its square.

anonymous · Answer

@amistre64 could you explain a little more what your doing up there?

amistre64 · Answer

just working some thoughts ...
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spose we have an even number, it is divisible by 2

if we multiply it by itself, we have 2 factors, each divisible by 2
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spose we have an odd number, it is NOT divisible by 2

if we multiply it by itself, we have 2 factors, NEITHER are divisible by 2
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spose we have a number that is the product of an even and an odd, it is divisible by 2

if we multiply it by itself, we have 2 factors that are still divisible by 2
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ergo

if a number is divisible by 2, its square is divisible by 2
if a squared number is divisible by 2, then its primary value is also divisible by 2

this creates a bidirectional setup

amistre64 · Answer

2n(2n) = 2(2n^2)

2n(2m+1) = 2(2nmk+n) = 2k, an even number

odd number are not divisible by 2, so any combination of odd numbers will not be divisible by 2 to start with.

amistre64 · Answer

i) 2 | a ⇒ 2| a^2 
if a is even, then a^2 is even .... so true

ii) 2 | a^2 ⇒ 2 | a 
since a^2 is even only if a is even ... this is also true

iii) 2 | a ⇔ 2 | a^2

since i and ii are true, iii is correct by default.