# 数学代写|数论作业代写number theory代考|MAST90136

## 数学代写|数论作业代写number theory代考|Waring’s Problem

Once again, having studied an idea involving perfect squares, it is natural to ask what happens with cubes, fourth powers, and so on. In 1770 Edward Waring (1736 – 1798) asked whether the four square theorem can be extended to higher powers. He conjectured that nine cubes would suffice (i.e. every positive integer can be written as a sum of nine cubes), and that 19 fourth powers would suffice. David Hilbert (1862 – 1943) proved that for each positive integer $k$, there is a number $g(k)$ such that every positive integer can be written as a sum of $g(k) k$-th powers, and no number smaller than $g(k)$ will suffice. We know that $g(1)=1$ since any positive integer can be written as itself, and we proved in the previous section that $g(2)=4$. It is now also known that $g(3)=9, g(4)=$ $19, g(5)=37$, and $g(6)=73$, so Waring was indeed correct in his conjectures.

Is there a general formula for $g(k)$ ? The following formula has been conjectured but not as yet proved.

Conjecture 9.13.
$$g(k)=2^k+\left[\left(\frac{3}{2}\right)^k\right]-2,$$
for all positive integers $k$, where $[x]$ denotes the greatest integer function.

Example 9.7. Lel’s subslitule a few values into the conjectured formula for $g(k)$ and see if they match up with the values which have been proven:
For $k=1,2^1+\left[(3 / 2)^1\right]-2=2+1-2=1$,
for $k=2,2^2+\left[(3 / 2)^2\right]-2=4+[9 / 4]-2=4+2-2=4$,
for $k=3,2^3+\left[(3 / 2)^3\right]-2=8+[27 / 8]-2=8+3-2=9$,
for $k=4,2^4+\left[(3 / 2)^4\right]-2=16+[81 / 16]-2=16+5-2=19$,
for $k=5,2^5+\left[(3 / 2)^5\right]-2=32+[243 / 32]-2=32+7-2=37$,
and
$$\text { for } k=6,2^6+\left[(3 / 2)^6\right]-2=64+[729 / 64]-2=64+11-2=73 \text {. }$$
This is remarkable evidence that the conjecture is true, but it is not, of course, proof that it is true for all positive integers $k$.

## 数学代写|数论作业代写number theory代考|Solved Problems

9.1. Find all solutions of the linear equation $10 x-6 y=17$.
Solution:
Since $\operatorname{gcd}(10,6)=2$ and 2 does not divide 17 , there are no solutions.
9.2. Find all solutions of the linear equation $10 x-7 y=17$.
Solution:
Since $\operatorname{gcd}(10,7)=1$, there are solutions. By inspection, $x=1$ and $y=-1$ is a solution. By Step (4) of our procedure in Section 9.2, the general solution is $(1+(-7) t,-1-10 t)=(1-7 t,-1-10 t)$. For example, if $t=1$, we have the solution $(-6,-11)$, and we can check that $10(-6)-7(-11)=-60+77=17$.
9.3. Find all solutions of the linear equation $10 x-6 y=18$.
Solution:
Since $\operatorname{gcd}(10,6)=2$ and 2 divides 18 , we can simplify to the equivalent equation $5 x-3 y=9$, which has solutions since gcd $(5,3)=1$. Since a solution is not evident by inspection, we first solve the equation $5 x-3 y=1$, which by inspection has a solution $(2,3)$. Multiplying through by 9 , we get $5(9 \cdot 2)-3(9 \cdot 3)=9$, i.e., we have found the solution $(18,27)$. We can check: $5(18)-3(27)=90-81=$ 9 . Hence the general solution is $(18-3 t, 27-5 t)$. Checking one of these solutions, if, say, $t=2$, we get the solution $(12,17)$, and we check that $5(12)-3(17)=60-51=9$.
9.4. Find one solution of the linear equation $2 x_1+4 x_2-6 x_3+5 x_4=$ 10.
Solution:
Since the coefficients 2 and 5 of $x_1$ and $x_4$ are relatively prime, we assign random values to $x_2$ and $x_3$, say $x_2=x_3=1$ and now solve the equation $2 x_1+5 x_2=10-4(1)+(6) 1=12$. A solution here is $x_1=1$ and $x_4=2$, so our solution (one of many!) is $(1,1,1,2)$. We check: $2(1)+4(1)-6(1)+5(2)=10$.
9.5. Suppose that $a x+b y=c$ with $\operatorname{gcd}(a, b)=1$, and suppose there are two solutions, $\left(x_0, y_0\right)$ and $\left(x_1, y_1\right)$ with $x_1=1+x_0$. Prove that $b=\pm 1$.

# 数论代考

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