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# 数学代写|数论代写Number Theory代考|Gauss’s Lemma

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## 数学代写|数论代写Number Theory代考|Gauss’s Lemma

In 1808 , Gauss discovered another very effective criterion to tell whether a number is a quadratic residue for a prime $p$. This criterion, now called Gauss’s lemma, enabled him to give a far less complicated proof of the law of quadratic reciprocity than the one he had been able to give when he was eighteen.

Here is how Gauss’s lemma works. For a number $a$ relatively prime to $p$, reduce the numbers in the list
$a, 2 a, 3 a, \ldots, \frac{p-1}{2} a$
to their residues from $-\frac{p-1}{2}$ to $\frac{p-1}{2}$.
For example, if $p=13$, then, for $a=3$, the list
$3,2 \cdot 3,3 \cdot 3,4 \cdot 3,5 \cdot 3,6 \cdot 3$
becomes
$$3,6,-4,-1,2,5 \text {. }$$
Note that each of the six integers $1,2,3,4,5,6$ from 1 to $\frac{p-1}{2}$ occurs exactly once in this list with either a plus or a minus sign. Gauss then tells us to count the number of minus signs; in this case there are two, and so, letting $n=2$, we do the following computation (since $a=3$ ):
$$\left(\frac{3}{13}\right)=(-1)^n=(-1)^2=1$$
and we get 1 , as we should, since 3 is a quadratic residue of 13 .
Let’s try it for $a=5$, a quadratic nonresidue of 13 . The list
$5,2 \cdot 5,3 \cdot 5,4 \cdot 5,5 \cdot 5,6 \cdot 5$
becomes
$$5,-3,2,-6,-1,4$$

## 数学代写|数论代写Number Theory代考|Euler’s Conjecture

Euler, without benefit of proof, but relying on massive calculations, conjectured that the beautiful patterns we have seen for the quadratic nature of $a=2$ and $a=3$ in Theorems 8.3 and 8.4 also hold more generally for other values of $a$. The most important feature of his conjecture is that whether a number $a$ is a quadratic residue of $p$ does not really depend on the prime $p$ itself; rather, it depends only on the remainder of $p$ modulo $4 a$.

We saw in Theorem 8.3 that the quadratic nature of $a=2$ depends not on $p$ itself, but only on the remainder of $p$ modulo 8 , and we saw in Theorem 8.3 that the quadratic nature of $a=3$ again depends not on $p$, but only on the remainder of $p$ modulo 12. So too, then, by Euler’s conjecture, the quadratic nature of $a=5$ for a prime $p$ will depend not on $p$ itself, but only on the remainder of $p$ modulo 20 .

Therefore, since $9^2 \equiv 5(\bmod 19)$, we know that 5 is a quadratic residue of 19 , and so, by this conjecture of Euler’s, we would immediately see that 5 is also a quadratic residue for all primes $p$ congruent to 19 modulo 20, that is, for $p=59,79,139,179,199,239, \ldots$.

We saw earlier that 5 is a quadratic nonresidue of 13 . Therefore, by this same conjecture of Euler’s, 5 should also be a quadratic nonresidue for the primes $p=53,73,113,173,193,233, \ldots$.

The other important feature of Euler’s conjecture involves the symmetry that is so evident in Theorems 8.3 and 8.4. For a number $a$, the quadratic nature of $a$ for a prime $p$ with remainder $r$ modulo $4 a$ is the same as it is for a prime $q$ with remainder $4 a-r$. For example, for $a=2$,if $p$ has remainder 1 modulo 8 , and $q$ has remainder $7=8-1$, then 2 is a quadratic residue of both $p$ and $q$; on the other hand, for $a=3$, if $p$ has remainder 5 modulo 12 , and $q$ has remainder $7=12-5$, then 3 is a quadratic nonresidue of both $p$ and $q$.

## 数学代写|数论代写Number Theory代考|Gauss’s Lemma

1808年，高斯发现了另一个判别一个数是否是素数的二次余数的有效准则$p$。这个标准，现在被称为高斯引理，使他能够给出一个比他18岁时给出的二次互易定律简单得多的证明。

$a, 2 a, 3 a, \ldots, \frac{p-1}{2} a$

$3,2 \cdot 3,3 \cdot 3,4 \cdot 3,5 \cdot 3,6 \cdot 3$

$$3,6,-4,-1,2,5 \text {. }$$

$$\left(\frac{3}{13}\right)=(-1)^n=(-1)^2=1$$

$5,2 \cdot 5,3 \cdot 5,4 \cdot 5,5 \cdot 5,6 \cdot 5$

$$5,-3,2,-6,-1,4$$

## MATLAB代写

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