Which integers can be expressed as $$a^3+b^3+c^3-3abc$$? $$a$$, $$b$$, $$c\in\Bbb Z$$.

$(a\pm1)^3+a^3+a^3-3(a\pm1)a^2=3a\pm1$

$(a-1)^3+a^3+(a+1)^3-3a(a+1)(a-1)=9a$

$2(a^3+b^3+c^3-3abc)=3(a+b+c)(a^2+b^2+c^2)-(a+b+c)^3$

If $$3\mid(a^3+b^3+c^3-3abc)$$, then $$3\mid(a+b+c)^3$$, $$3\mid(a+b+c)$$. so $$9\mid(a^3+b^3+c^3-3abc)$$.

All $$n$$ such that $$3\nmid n$$ or $$9\mid n$$.

1.21 唯一因式分解定理的证明

2.5  五边形与五角数

$p_m(k)=\dfrac{mk(k-1)}2+k.$

2.8  一个不平凡的结论

2.9 什么数恰好有 $$60$$ 个因数?

$$kn = x^2+y^2+1$$

$$n$$ is a odd number, then there exists positive integer $$k\gt0$$ such that $$kn = x^2+y^2+1$$ for some integers $$x,y$$.

with use of the Chinese remainder theorem we have to solve this problem only for power of primes:

suppose that $$n=p_1^{a_1}p_2^{a_2}\dotsm p_k^{a_k}$$, then we know that for each $$i$$, there exist $$x_i, y_i$$ such that $$p_i^{a_i}$$ divides  $$x_i^2+y_i^2+1$$. Now consider these equations:

$X\equiv x_i\pmod {p_i^{a_i}}, i= 1,2,\dotsc,k.$

these equations have solution because of  Chinese remainder theorem.

similarly these equation have solution:

$Y\equiv y_i\pmod {p_i^{a_i}}, i= 1,2,\dotsc,k.$

now $$n$$ divides  $$X^2+Y^2+1.$$

then we can apply hansel’s lemma. Actually we want to show that if for some $$\alpha$$, there exist $$x,y$$ such that $$p^\alpha$$ divides  $$x^2+y^2+1$$, then for  $$\alpha +1$$ such $$x$$  and $$y$$ exist. For this because in case $$\alpha$$, $$p$$ cannot divide both $$x$$  and $$y$$, then we can use hansel for improve $$\alpha$$ to $$\alpha+1.$$

#### References

1. 华罗庚, 数论导引.
2. Hardy, An introducton to the theory of numbers. 有中文本
3. Tom M. Apostol, Introduction to analytic number therory. 有中文本