# For pairwise coprime $a_1<\cdots<a_n$ and $b_i=((a_1\cdots a_n)/a_i)^{-1} \text{ mod } a_i$, $0>b_i>-a_i$, when $b=-1/a + \sum_{i=1}^n (b_i/a_i)=-1$?

Let $$2\leq a_1 be pairwise coprime integers, let $$a=a_1 \cdots a_n$$, and let $$b_i$$ be the unique integer such that $$b_i=(a/a_i)^{-1} \text{ mod } a_i$$ and $$0>b_i>-a_i$$. Then let $$b=-1/a + \sum_{i=1}^n (b_i/a_i)$$. It is known that $$b$$ is a negative integer such that $$-n.

Actually I am studying Saveliev's book Invariants for homology 3-spheres, and the integer $$b$$ arises here in the following way: The Seifert-fibered homology sphere $$\Sigma(a_1,\dots,a_n)$$ (defined in chapter 1.1.4) bounds a unique negative definite plumbing 4-manifold whose graph is a star-shaped tree, and the weight of the central node is exactly $$b$$.

Consider the case $$n=4$$, so in this case $$b=-1, -2$$ or $$-3$$. Is there a way to find a condition for $$a_1,a_2,a_3,a_4$$ so that $$b=-1$$? (Or at least an example.) I computed $$b$$ for some small $$a_i$$'s; for example if $$(a_1,a_2,a_3,a_4)=(2,3,5,7)$$ then $$b=-2$$, and if $$(a_1,a_2,a_3,a_4)=(2,5,7,11)$$ then $$b=-2$$, and I couldn't find $$(a_1,\dots,a_4)$$ with $$b=-1$$. I am wondering to find a criterion for $$(a_1,\dots,a_4)$$ to have $$b=-1$$ by using some number theory, etc.

• Let $\,x\,$ be the solution of $\,x\equiv 1\pmod{\!a_i},\, i=1,\cdots, n\,$ that is given by the well-known CRT formula, so $\,x = 1+a\:\!b,\,$ for some $\,b\in \Bbb Z.\,$ Your $\,b\,$ is the "CRT quotient" $\, b = (x\!-\!1)/a\ \$ Commented Jul 25, 2022 at 15:09
• @BillDubuque Yes but $x$ is determined only modulo $\prod_i a_i$ Commented Jul 28, 2022 at 12:47

For $$b=-1$$:

$$n\ \ (a_1,a_2,...,a_n)$$

$$3\ \ (2, 3, 7)\ (2, 5, 7)\ (3, 4, 5)\ (4, 5, 7)\ (5, 6, 7)\ ...$$

$$4\ \ (2, 5, 9, 13)\ (3, 4, 7, 11)\ (3, 5, 8, 11)\ (5, 8, 9, 11)\ ...$$

$$5\ \ (3, 4, 11, 13, 19)\ (5, 7, 13, 16, 19)\ (5, 11, 14, 17, 19)\ ...$$

$$6\ \ (5, 9, 11, 13, 16, 17)\ (4, 7, 13, 17, 19, 23)\ (7, 9, 10, 13, 17, 23)\ ...$$

• Thanks. Did you find these examples by hand? Commented Aug 16, 2022 at 6:20
• I find them by computer. I haven't found any criterion yet. Commented Aug 17, 2022 at 1:09