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21 votes

Is there a generalization of factoring that can be extended to the Real numbers?

This is a question that is usually answered via Ring Theory. You can think of a ring as a set where addition, substraction, and multiplication is well defined. Here, we aren't thinking of ...
Daniel5803's user avatar
6 votes

Is there a generalization of factoring that can be extended to the Real numbers?

Obviously you can factor a real number into any set of numbers whose product is that real number. As for PRIME factorizations I know of one generalization of that probably DOESN'T work well for real ...
Mr. Nichan's user avatar
4 votes
Accepted

Proof Verification: If $p^k m^2$ is an odd perfect number with special prime $p$, then $p \equiv 1 \pmod 8$ holds.

Given an odd perfect number of the form $p^km^2$ with $p\equiv k\equiv1\pmod{4}$, you write $$m^2-p^k=a^2-b^2\qquad\text{ and }\qquad m^2-p^k=2^rt,$$ where $a$, $b$, $r$ and $t$ are positive integers. ...
Servaes's user avatar
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3 votes

Is there a generalization of factoring that can be extended to the Real numbers?

Let's take a different perspective than the other answers have taken so far. You can think of prime factorizations as telling you exactly how the natural numbers $\mathbb{N}$ (here I am excluding zero)...
Qiaochu Yuan's user avatar
2 votes

ACT practice test, aren't both $3$ and $12$ viable answers?

In order for there to be two distinct real-valued solutions to $ax^2 + bx + c = 5x^2 + 16x + c = 0$, the discriminant $b^2 - 4ac$ must be strictly positive; hence we require $$16^2 - 4(5)(c) > 0,$$ ...
heropup's user avatar
  • 141k
1 vote

Is there a generalization of factoring that can be extended to the Real numbers?

Any rational $x>0$ can be uniquely written as $x=\prod{p_i}^{m_i}$ with $p_i$ the $i^\text{th}$ prime and $m_i\in\mathbb Z$ its associated multiplicity. This extends the notion of factorization to ...
fgrieu's user avatar
  • 1,768
1 vote

Factoring $N = pq$ for primes $p,q$ knowing $q\bmod (p-1)$

Hint: $\bmod p\!:\ a^N\equiv a^{p(k+j(p-1))} \equiv a^{k+j(p-1)}\equiv a^k(a^{p-1})^j \equiv a^k,\,$ hence $\,p\mid \gcd(N,\color{#c00}{a^{N-k}-1})\,$ if $\,p\nmid a.\,$ The gcd is quickly computable ...
Bill Dubuque's user avatar

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