- #1

Firepanda

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So far for p

^{2}= <2> I have:

Show p

^{2}is containd in <2>

Since [we have already shown]

p={r + s√-5 | r = s mod 2}

Then p

^{2}elements are of the form (r+s√-5)

^{2}

= (r

^{2}-5s

^{2}) + 2rs√-5

Since r&s have the same sign, then (r

^{2}-5s

^{2}) is always even, & 2rs is always even too

=> p

^{2}elements are contained in <2>

Now, show <2> is contained in p

^{2}

=> show 2(a+b√-5) is an element of p

^{2}

so show 2a + 2b√-5 is an element of p

^{2}

From the above we know elements of p

^{2}take the form of 2x + 2y√-5

So <2> is contained in p

^{2}

Am i done for this part?

Thanks