Sum of three cubes: what is '4 or 5 mod 9'?
Does anyone know how to do Modulo operations in this Calculator?
Add or subtract the following in the given modulus: a. 5 mod 9 + 8 mod 9 . 5 mod 9- 8 mod 9 - 25 mod 30 + 5 mod 30 . 12 mod 15 - 6 mod 15 2. 6 mod 10 + 6 mod 10
elementary number theory - Help with congruences and positive divisors - Mathematics Stack Exchange
How much is 17 mod 3?
17 mod 3 equals 2 since dividing 17 by 3 gives a quotient of 5 and a remainder of 2. The remainder is the result of the modulus operation. In simpler terms, 17 mod 3 = 2.
How to calculate modulo division?
To calculate modulo division: subtract the divisor from the dividend until the resultant is less than the divisor.
What are the components of modulo division?
The components of modulo division are dividend, divisor, quotient, and remainder. The remainder is the answer or end result of the operation.
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Hello all,
All articles about sums of three cubes have this, and I don't understand it. The rough idea is that all numbers n can be created by summing three cubes, unless n is 4 or 5 mod 9. What is meant by 4 or 5 mod 9? I have a basic knowledge of what modulo is, and 4 or 5 mod 9 gives 4 or 5. Is mod used here to describe a set of numbers? If so, how so?
Thanks for reading.
Your solution is correct. Below, I've provided some further explanation as to how exactly it works.
Say we want to find all $\alpha\in\mathbb{N}$ with $5^\alpha\equiv 2\pmod 9$. Given any two $\alpha,\beta\in\mathbb{N}$ with $5^\alpha\equiv 2\pmod 9$ and $5^{\alpha+\beta}\equiv2\pmod 9$,we have $5^\beta\equiv 1\pmod 9$. On the other hand, if given some $\alpha,\beta\in\Bbb{N}$ with $5^\alpha\equiv 2\pmod 9$ and $5^{\beta}\equiv 1\pmod 9$, we have $5^{\alpha+\beta}\equiv 2\pmod 9$.
Also, if we have two $\alpha,\beta$ with $5^\alpha\equiv 1\pmod 9$ and $5^{\beta}\equiv 1\pmod 9$, we would have $5^{\gcd(\alpha,\beta)}\equiv 1\pmod 9$. Therefore there exists some $d\in\mathbb{N}$ such that $5^\alpha\equiv 1\pmod 9$ if and only if $d\mid \alpha$.
Combining these results gives that $5^\alpha\equiv 2\pmod 9$ if and only if $\alpha \equiv \beta\pmod d$ for some $\beta$ with $5^\beta\equiv 2\pmod 9$.
We find that when working modulo $9$, we can take $d=6$ (simply the smallest positive integer $d$ such that $5^d\equiv 1\pmod 9$) and an example of a solution would be $5$. Putting $d=6$ and $\beta=5$, this gives: $$\alpha\equiv 5\pmod 6$$ You can do the same for the other congruence.
Usin the Chinese Remainder theorem, we have to solve for \begin{cases}5^n\equiv 2\mod9,\\5^n\equiv 3\mod11.\end{cases} You won't have to examine so many cases, since by Euler's theorem, $5$ has order a divisor of $\varphi(9)=6$ modulo $9$, and a divisor of $\varphi(11)=10$ modulo $11$.
Calculating the successive powers of $x$ mod. $9$ and mod. $11$, we see that $5$ has indeed order $6$ mod $9$, but order $5$ mod $11$, so you only have to calculate $30$ modular powers of $5$, and as you found the values $2$ and $5$, you obtain $$\begin{cases} 5^n\equiv2\mod 9\phantom{1}\iff n\equiv 5\mod6,\\5^n\equiv 3\mod 11\iff n\equiv 2\mod 5.\end{cases}$$ The solutions result from a Bézout's relation between $6$ and $5$, especially simple here: $\;6-5=1$, so the solutions satisfy $$n\equiv 2\cdot 6-5\cdot 5=-13\mod \operatorname{lcm}(6,5)=30.$$ The smallest positive value is $17$, and the solutions at most equal to $140$ are $$\{17,47,77,107,137\}.$$