Primitive roots: If $p$ is a prime such that $p\equiv 1 \pmod 4$, and $a$ is a primitive root, then $-a$ is also a primitive root. [duplicate]

Primitive roots: If $p$ is a prime such that $p\equiv 1 \pmod 4$, and $a$ is a primitive root, then $-a$ is also a primitive root.

In this particular question I did show that in fact $(-a)^{p-1} \equiv 1 \pmod 4$, what is trivial. But, how can I assure that $p-1$ is its order?

Thanks a lot.

marked as duplicate by lab bhattacharjee algebra-precalculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 31 '15 at 17:02

• You mean : If $a$ is a primitive root,$-a$ is one as well. – Peter Oct 31 '15 at 16:14
• There seems to be some confusion in the Question. Apart from the omission of an assumption about $a$ being "a primitive root", as Peter noted, it seems unlikely that being a primitive root mod 4 is the real issue. – hardmath Oct 31 '15 at 16:18
• The edit to the title is helpful, but the same change in the body is more important. Also, please spell out in the body that $a$ is a primitive root mod $p$. – hardmath Oct 31 '15 at 16:25

We are given that $a$ has order $p-1$ modulo $p$, and want to show that $-a$ also has order $p-1$. Suppose to the contrary that $(-a)^n\equiv 1\pmod{p}$ for some $n\lt p-1$. Then $(-a)^{2n}\equiv 1\pmod{p}$, so $a^{2n}\equiv 1\pmod{p}$. Since $a$ is a primitive root of $p$, it follows that $n=\frac{p-1}{2}$.

We show that this is impossible. Recall that since $p\equiv 1\pmod{4}$, there is an $x$ such that $x^2\equiv -1\pmod{p}$. Then $$1\equiv (-a)^{(p-1)/2}\equiv(x^2a)^{(p-1)/2}\equiv a^{(p-1)/2}\pmod{p},$$ contradicting the fact that $a$ has order $p-1$.

Assume, $p-1$ is the order of $a$ modulo $p$ and assume $(-a)^m \equiv 1 \pmod p$ for some $m<p-1$. Then $m$ must be odd because otherwise we would have $a^m\equiv 1 \pmod p$, which is impossible.

For such an odd $m$, we have $a^m\equiv -1\pmod p$ and therefore $a^{2m}\equiv 1\pmod p$. But $2m=p-1$ is impossible because $p-1$ is divisible by $4$. So, we have a contradiction again.

If $p\equiv 1 \pmod 4$, it means that $p-1\equiv 0 \pmod 4$.

Therefore $(-a)^{p-1}\equiv (-a)^{0} \equiv 1 \pmod 4$.

I would do it in this way.

• Aren't you just repeating what the OP says already in the body of the Question? – hardmath Oct 31 '15 at 16:21
• I'm afraid, I can have misunderstood the question. – Michael Oct 31 '15 at 16:23