- #1
Firepanda
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I'm doing part iii) here:
So far for p2 = <2> I have:
Show p2 is containd in <2>
Since [we have already shown]
p={r + s√-5 | r = s mod 2}
Then p2 elements are of the form (r+s√-5)2
= (r2 -5s2) + 2rs√-5
Since r&s have the same sign, then (r2 -5s2) is always even, & 2rs is always even too
=> p2 elements are contained in <2>
Now, show <2> is contained in p2
=> show 2(a+b√-5) is an element of p2
so show 2a + 2b√-5 is an element of p2
From the above we know elements of p2 take the form of 2x + 2y√-5
So <2> is contained in p2
Am i done for this part?
Thanks
So far for p2 = <2> I have:
Show p2 is containd in <2>
Since [we have already shown]
p={r + s√-5 | r = s mod 2}
Then p2 elements are of the form (r+s√-5)2
= (r2 -5s2) + 2rs√-5
Since r&s have the same sign, then (r2 -5s2) is always even, & 2rs is always even too
=> p2 elements are contained in <2>
Now, show <2> is contained in p2
=> show 2(a+b√-5) is an element of p2
so show 2a + 2b√-5 is an element of p2
From the above we know elements of p2 take the form of 2x + 2y√-5
So <2> is contained in p2
Am i done for this part?
Thanks