Where is the pattern?

I found $\phi(720)=192$ , and since $P$ is the product of all of those 192 relatively prime numbers to 720, P must also be relatively prime to $720$ since they share no common factors. Then using Euler's theorem, $P^{192}\equiv1\pmod{720}$ , and then it's easy to see that $(P^{2})^{96}\equiv1\pmod{720}$ or ${P^{2}\equiv1\pmod{720}}$ hence the remainder is $\boxed1$ .

Edit:

The modular inverses approach is most accurate.. Since there is a slight inaccuracy in the above solution (taking the $96th$ root on both sides of the modular expression can yield solutions other than $1$ , although there aren't any) I'm trying an alternative approach using the Chinese Remainder Theorem and Carmichael function. Please feel free to correct me. I'm not sure if this is correct..

The Carmichael function ${\lambda(n)}$ states that $a^{\lambda(n)}\equiv1\pmod{n}$ when $a$ and $n$ are coprime and $a<n$ . Let $a_1, a_2, a_3, \ldots a_n$ be all of the numbers which are relatively prime to 720. We get congruences such as

• $a_1^{\lambda{(720)}}\equiv1\pmod{720}$
• $a_2^{\lambda{(720)}}\equiv1\pmod{720}$
• $a_3^{\lambda{(720)}}\equiv1\pmod{720}$ ...
• $a_n^{\lambda{(720)}}\equiv1\pmod{720}$

and then we can multiply all of the separate modular congruences to obtain

$(a_1^{12} a_2^{12} a_3^{12}\ldots a_n^{12})=P^{12}\equiv1\pmod{720}$ .

Now, we need to scale down from $P^{12}$ to $P^{2}$ . To do this, we can see that $720=16\times9\times5$ and ${\lambda(16)}=4$ , ${\lambda(9)}=6$ , ${\lambda(5)}=4$ .

We can write the congruences as

• $P^{\lambda(16)}\equiv1\pmod{16}$ which gives $P^{4}\equiv1\pmod{16}$
• $P^{\lambda(9)}\equiv1\pmod{9}$ which gives $P^{6}\equiv1\pmod{9}$
• $P^{\lambda(5)}\equiv1\pmod{5}$ which gives $P^{4}\equiv1\pmod{5}$

As P is the product of all integers relatively prime to 720, P must also be coprime to 16, 9 and 5.

The expressions, upon solving for the residues of $P^{2}$ , yield

• $P^{2}\equiv1, 7, 9, 15\pmod{16}$
• $P^{2}\equiv1, 4, 7\pmod{9}$
• $P^{2}\equiv1, 4\pmod{5}$ .

Now we see that for $P$ to exist, $P^{2}$ has to be congruent to $1$ for all three modulos, since if $P^{2}$ is congruent to say, for example, 9 for the first, 4 for the second and 1 for the third, upon solving we would obtain no value of $P$ satisfying all three congruences.

• $P^{2}\equiv1\pmod{16}$
• $P^{2}\equiv1\pmod{9}$
• $P^{2}\equiv1\pmod{5}$

Then after solving these three congruences simultaneously, we obtain the final answer of ${P^{2}\equiv1\pmod{720}}$ .

${ \left( \prod _{ a\in P }^{ \quad }{ a } \right) }^{ 2 } \pmod{n} \equiv 1$

Yes, it's true for any mod! This is because elements of $P$ may be classified in two ways:

(1) $a^2=1$ . These elements are their own multiplicative inverses, and when the product is squared the do not contribute anything.

(2) $ab=1$ . If $a$ has a multiplicative inverse, it is necessarily in $P$ . Otherwise, $b$ would have some factor of $n$ in it and could only be a multiple of that when multiplied by another element, thus the product could not one. In addition, if $a$ has a multiplicative inverse in $P$ is it unique to $a$ .

Consider the existence of another multiplicative inverse to $a$ , $c$ . Then by writing down multiples of $a$ we run into two ones. These two ones ought to be the same distance from a 0: equality is preserved going backwards, and we know that if $b<c$ then by going backwards from both $b$ times we reach zero for both. Yet there exists only one zero at $a\times0$ because $a\in P$ . Thus $b=c$ .

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Okeydoke, now that that's all done we can just pair multiplicative inverses to get all 1's for those and squaring 1 gives 1, and then by adding all $a$ with $a^2=1$ we still have 1 for the product.

here is the solution of the problem in a image bello w

For any integer n in P, there exists a unique integer m also in P such that nm ≡ 1 (mod 720). (note that n and m could be equal).

Since there are an even number of integers being multiplied in the product P², we can group the integers in pairs such that every pair has a product of 1 (mod 720), and the product of all pairs is therefore $\boxed{1}$ .

Since the polynomial $(x-a_1)(x-a_2)...(x-a_{\varphi(720)})\equiv x^{\varphi(720)}-1\pmod{720}$ , where a_i is a reduced residue system, we can plug in x=0.