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September 19

Sets where lots of numbers can be added to themselves or multiplied by themselves with the same result

In the familiar set of natural numbers, we only have 2. But in a set like Z8, we have 2 and 6. In Z8, 6+6 and 6*6 are both 4. How about in any sets from Z9 to Z14?? Z15 has 5; in Z15 5+5 and 5*5 are both 10. Georgia guy (talk) 16:03, 19 September 2016 (UTC)[reply]

It seems that your question is to find a semigroup where the equation $x^{2}=2x$ has "many" solutions. Well, that is still a very open question. In the case of a group (i.e. every element has an inverse) then the equation is satisfied iff $x=2\lor x=0$ (if both $x-2$ and $x$ have an inverse, then $x^{2}-2x$ also has one), which means there will be few solutions (well, 2 exactly).

So for starters, it will not have many solutions in Zp with p prime (p is prime iff. Zp is a group). Tigraan^{Click here to contact me} 16:23, 19 September 2016 (UTC)[reply]

How about solutions with Zk where k is any natural number?? I gave 2 solutions already. Any pattern in where solutions occur?? Georgia guy (talk) 16:28, 19 September 2016 (UTC)[reply]
The Z15 case is the beginning of a group of solutions, not sure there are others. Let p & q be twin primes, so q=p+2. Then in Z(p*q) q+q = q*q. So in Z35, 7+7=7*7 and in Z143, 13+13=13*13. In fact this works for any two numbers that are two apart, but both the multiplication may lead zero. In fact in Z8, 4 also qualifies, since 4+4 = 4*4=0, making them twin primes just makes sure the sum/product is not zero.Naraht (talk) 18:49, 19 September 2016 (UTC)[reply]
Correction: p and q don't have to be prime. In Z24, 6+6 and 6*6 are both 12. Georgia guy (talk) 19:28, 19 September 2016 (UTC)[reply]

For modular arithmetic

Z_{p}

, you are asking for solutions of the form

x^{2}-2x-kp=0

, where

k=0,1,2,...

The quadratic formula gives

x=1+{\sqrt {1+kp}}

, where

{\sqrt {1+kp}}

is an integer and x < p. Thus k=0 gives the the solution x=2. For instance, p=8, k=3, yields x=6 and so 6+6=6*6 mod 8. But also p=12, k=2 and p=24, k=1 are also 6+6=6*6 solutions. --Mark viking (talk) 20:49, 19 September 2016 (UTC)[reply]

In the case of commutative rings where 2 is invertible, then $x^{2}=2x$ means y= x/2 is an idempotent, so this amounts to finding idempotents. As in Decompose your ring using idempotents example 2, each idempotent corresponds to breaking the spectrum of the ring into two open sets, a complete set of orthogonal idempotents correspond to writing the ring as a direct product or the spectrum as a union of connected components, and any idempotent is a sum of these orthogonal ones. In the case of $Z_{n}$ , you can write it as a product of the $Z_{p}m$ from the factorization into prime powers p^m and the Chinese remainder theorem. So this gives all the idempotents & the solutions of $x^{2}=2x$ in the case n is odd.John Z (talk) 05:43, 20 September 2016 (UTC)[reply]