Verified answer

1) The only idempotents in Z/<p^k> are 0 and 1.

Proof: Suppose that some nonzero x is a solution of x^2 = x (mod p^k).

Writing x = p^r * s for some integer r in {0, ..., k-1} with gcd(p, s) = 1, we have by substitution

(p^r * s)^2 = p^r * s (mod p^k)

==> p^(2r) * s^2 - p^r * s = 0 (mod p^k)

==> p^r * s * (p^r * s - 1) = 0 (mod p^k)

==> s * (p^r * s - 1) = 0 (mod p^(k-r))

==> p^r * s - 1 = 0 (mod p^(k-r)), since gcd(p, s) = 1

==> p^r * s = 1 (mod p^(k-r)).

This can only be true if r = 0 and hence x = s = 1 (mod p^k).

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2) Note that 12 = 2^2 * 3.

The idempotents in Z/<2^2> are 0 and 1, and the idempotents in Z/<3> are 0 and 1.

By the Chinese Remainder Theorem, we have four possible idempotents in Z/<12>:

(i) x = 0 (mod 2^2) and x = 0 (mod 3) ==> x = 0 (mod 12)

(i) x = 0 (mod 2^2) and x = 1 (mod 3) ==> x = 4 (mod 12)

(i) x = 1 (mod 2^2) and x = 0 (mod 3) ==> x = 9 (mod 12)

(i) x = 1 (mod 2^2) and x = 1 (mod 3) ==> x = 1 (mod 12).

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3) This is along the lines of problem #2.

By the Chinese Remainder Theorem, Z/<n> = Π(i=1 to n) Z/<pi^(ni)>.

Since 0 and 1 are the only idempotents in Z/<pi^(ni)> for each i (by problem #1), an idempotent in Z/<n> satisfies the system of congruences x = ai (mod pi^ni) for each i = 1, ..., n with ai = 0 or 1.

Since each ai has 2 choices and there are n congruences in the system,

we have 2^n possible systems, with 2^n solutions in total (via Chinese Remainder Theorem).

I hope this helps!

on a similar time as engaged on it, for 2 circumstances i seen this: because p^4-a million is even, 2 is a ingredient, and 2 is the smallest integer ingredient. for this reason (p^4-a million)/2 is the wonderful integer quantity that divides p^4-a million. even with the undeniable fact that, this looks an unusual answer to me because of the fact it has virtually not something to do with top p > 5 and ability of four, different than that p is unusual because of the fact it fairly is a significant and not 2. What did I pass over out? Edit: I remembered what I neglected out: a single quantity is mandatory that divides p^4-a million for each top p>5. So forgetful :( below is what I had tried. p^4-a million = (p-a million)(p+a million)(p^2+a million) because p>5, permit q = p-5, a favorable even quantity, p^2 + a million = q^2 + 10q +25 + a million = q^2 + 10q +26, which has no genuine roots = 4r^2 + 20r + 2*13, the place 2r = q = 2(2r^2 + 10r + 13) does not have integer components different than 2. for this reason, p^4-a million = (p-a million)(p+a million)(p^2+a million) = (p-a million)(p+a million)2(2r^2 + 10r + 13) Ooops, 2d time i seen the comparable concern that i discussed on the initiating.