# Solution to $\sum (-1)^k \binom{n}{k} \alpha_k = b_n$? [closed]

Is there anyone can tell me any information about the integer solution to the combinatotial equation $$\sum (-1)^k \binom{n}{k} \alpha_k = b_n$$ (all variables are integers)? For example, suppose alpha_{0}=0 , when n=2, it is 2alpha_{1}-alpha_{1}=-b_{2} If we take b_{2} to be a given number, this is a first degree Diophantine Equation, we know how to solve it using elementary number theory, right? But when n=3, take b_{3} to be a given number, alpha_{0}=0, could you write down a general solution to this Dioph. equation? Thank you very much!

## closed as unclear what you're asking by Ben McKay, Henry.L, RP_, Stefan Waldmann, Mikhail KatzSep 7 '17 at 14:31

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• What exactly do you want? {n, 1} is 1, so alpha_1 is always uniquely determined by the other alpha_i. – darij grinberg Apr 11 '10 at 23:36
• a is a fixed integer and this is an equation about alpha_{k}? – mingming Apr 11 '10 at 23:40
• Yes, so you can choose values of the other alpha_k, k > 0 arbitrarily and this determines the value of alpha_0 (which I'm sure is what darij meant to say). Perhaps you meant to ask about a sequence a_n on the right? – Qiaochu Yuan Apr 11 '10 at 23:49
• Can you write down a general solution to this Diophantine Equation? – mingming Apr 12 '10 at 0:03
• mingming, I hope you can see that this is not the question you originally asked. – Qiaochu Yuan Apr 12 '10 at 1:22

## 1 Answer

I figured that's the question you wanted to ask. The relation

$$\sum_{k=0}^{n} (-1)^k {n \choose k} a_k = b_n$$

for all $n$ (you did not specify this; it was very unclear) is equivalent to the relation

$$e^{x} A(-x) = B(x)$$

where $A(x) = \sum_{k \ge 0} \frac{a_k}{k!} x^k, B(x) = \sum_{k \ge 0} \frac{b_k}{k!} x^k$. This gives $A(x) = e^x B(-x)$, or

$$\sum_{k=0}^{n} (-1)^k {n \choose k} b_k = a_n.$$

So the $a_i$ are all integers if and only if the $b_i$ are all integers, and each uniquely determines the other. I don't know what else to say; you can choose either the $a_i$ or the $b_i$ arbitrarily. What exactly do you want to know?

• Thank you very much for your answer! Suppose alpha_{0}=0 , when n=2, it is 2alpha_{1}-alpha_{1}=-b_{2} If we take b_{2} to be a given number, this is a first degree Diophantine Equation, we know how to solve it using elementary number theory, right? But when n=3, take b_{3} to be a given number, alpha_{0}=0, could you write down a general solution to this Dioph. equation? – mingming Apr 12 '10 at 1:51
• You can use elementary number theory method to solve this three variable Dioph. equation. But when n turn out to be 4,5,6, how can we write down a general solution? – mingming Apr 12 '10 at 1:54
• I hope the reciprocity formular will work here for bigger n. – mingming Apr 12 '10 at 1:56
• Once more, mingming, that is not the question you originally asked. My understanding of the question you are currently asking is that you can write down a general solution by repeated use of Bezout's lemma. – Qiaochu Yuan Apr 12 '10 at 4:10
• Dear Mr. Yuan: Thank you very much. Could you use Bezout Lemma to write down the solution for n=3? I want to check it with my original solution. Best! – mingming Apr 12 '10 at 4:14