On a conjecture that $\sum_{i=1}\limits^na_ip_i = 1\Rightarrow \exists\big(\sum_{i=1}\limits^na_i=0\big)$ (about primes)

Question

Conjecture:

Denote by $p_i$ the $i^{\text{th}}$ prime number; by $a_i$ the $i^\text{th}$ arbitrary integer; and by $\exists(x)$ the existence of a chosen $x$.$$\sum_{i=1}^na_ip_i = 1\Rightarrow \exists\bigg(\sum_{i=1}^na_i=0\bigg)\tag{$n>1$}$$ In words: if the sum of an arbitrary quantity of successive prime numbers each multiplied by an integer is equal to $1$, then there exists a solution for which the sum of strictly all such integers is equal to $0$.

Examples:

$$\begin{align} 2(-1) + 3(1) &= 1 \tag{$-1+1=0$} \\ 2(1)+3(-2)+5(1)&=1\tag{$1-2+1=0$} \\ 2(1)+3(-3)+5(3)+7(-1)&=1\tag{$1-3+3-1=0$} \\ 2(-1)+3(2)+5(0)+7(-2)+11(1)&=1\tag{$-1+2+0-2+1=0$}\\&\vdots\end{align}$$

Even better: I have potentially found a method to find such integer coefficients! It is difficult to explain, but I will try my best. Once you get it, it should be all-good.

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Consider the difference of the second and first prime: $3-2=1$ (always subtract from the bigger number, albeit this has already been done). Or, $$\underbrace{p_2}_{3}-\underbrace{p_1}_{2}=1.\tag1$$ This is all we did:

$3-2=1 = Eq. (1)$

Now, consider the difference of the third and second prime: $5-3=2$. But also consider the overall difference of: this result (which is $2$) and $3-2$ (which is $1$). This should give you $2-1$ which is $1$. Equivalently, we have: $$\underbrace{\underbrace{(p_3-p_2)}_{5\,-\,3\,=\,\color{red}2}-\underbrace{(p_2-p_1)}_{3\,-\,2\,=\,\color{red}1}}_{\color{red}2\,-\,\color{red}1\,=\,1}= p_3-p_2-p_2+p_1 = p_3-2p_2+p_1 = 1.\tag2$$ This is all we did:

$5-3=2$.
$2-(3-2)=1\Rightarrow (5-3)-(3-2)= Eq. (2)$

Thirdly, consider the difference of the fourth and third prime: $7-5=2$. But also consider the difference of: this result (which is $2$) and $5-3$ (which $2$). This should give you $2-2=0$. Now also consider the difference of: this result (which is $0$) and $p_3-2p_2+p_1$ (which is $1$). But remember to subtract from the biggest number, so instead of obtaining $0-1=-1$, we get $1-0=1$. Equivalently, we have: $$p_3-2p_2+p_1-\underbrace{\big(\underbrace{(p_4-p_3)}_{7\,-\,5\,=\,\color{red}2}-\underbrace{(p_3-p_2)}_{5\,-\,3\,=\,\color{red}2}\big)}_{\color{red}2\,-\,\color{red}2\,=\,0}=\cdots =p_1-3p_2+3p_3-p_4=1.\tag3$$

This is all we did:

$7-5=2$.
$2-(5-3)=0$.
$0 - (p_3-2p_2+p_1)=-1$ and then reversed subtraction because $p_3-2p_2+p_1 > 0$.
$(p_3-2p_2+p_1)-0=1\Rightarrow (p_3-2p_2+p_1) - (2-(5-3))=Eq. (3)$

Notice that all the equations $(1)$, $(2)$ and $(3)$ are respectively identical to the first three examples of my conjecture (but just rearranged and notated differently)! Each equation is equal to $1$, and the sum of strictly all the integer coefficients in each equation is each equal to $0$.

But did you notice the pattern to the method? Let's continue once more!

$11-7 = 4$.
$4 - (7-5) = 2$.
$2 - ((7-5)-(5-3))=2$.
$2-(p_1-3p_2+3p_3-p_4)=1\Rightarrow 2 - ((7-5)-(5-3)) - (p_1-3p_2+3p_3-p_4)=Eq. (4)$ below (after $p_i$ substitution).

$$\begin{align}&p_1 - \big((p_4-p_3)-(p_3-p_2)\big)-(p_1-3p_2+3p_3-p_4) \\ &=-p_1+2p_2+0p_3-2p_4+p_5\\&=1.\end{align}\tag4$$

It is clear that $Eq. (4)$ is the fourth example of my conjecture.

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If you understood the pattern of my method, great! If not, we can discuss it (hopefully in clearer detail) in a chat. I am young so my explanation skills might not be that great.

In fact, this method is not that great (either); it quickly becomes incredibly tedious. Firstly, it requires Gilbreath's Conjecture to be true (i.e. that $1$ has to always be there in my conjecture) and secondly, the algebra gets messy.

Does anyone have any comments on this? I mean, why does this method even (seem to) work?

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Thank you in advance, and apologies for a long post.

Answer 1

Your conjecture is true. Suppose that one set of solutions is given by $A_i$. As the equation $\sum\limits_{i=1}^n a_ip_i=1$ is Diophantine, one can show that the general solution is given by $a_1=A_1+\sum\limits_{r=1}^{n-1}t_rp_{r+1}$ and $a_k=A_k-p_1t_{k-1}$ where $t_r$ are some integers to be determined and $1<k\le n$. Now $$\sum\limits_{i=1}^n a_i=\sum_{i=1}^nA_i+\sum\limits_{r=1}^{n-1}t_rp_{r+1}-\sum_{r=1}^{n-1}p_1t_r=0$$ from the constraint. Since $\sum\limits_{i=1}^nA_i=0$, it is sufficient to solve $\sum\limits_{r=1}^{n-1}(p_{r+1}-p_1)t_r=0$ for $t_r$. This can be rewritten as $(p_2-p_1)t_1+\sum\limits_{r=2}^{n-1}(p_{r+1}-p_1)t_r=0\implies \sum\limits_{r=2}^{n-1}(p_{r+1}-p_1)t_r=2$ after letting $t_1=-2$ and noting that $p_2-p_1=3-2=1$. This new Diophantine equation has infinitely many solutions for $t_r$, where $1<r\le n$.

Answer 2

Assume that we are given

$$\sum_{i=1}^{n}a_ip_i=1$$

with $\sum_{i=1}^{n}a_i=M$.

Suppose we produce a solution of $$\sum_{i=1}^{n}x_ip_i=0$$ such that $\sum_{i=1}^{n}x_i=-M$. Then replacing $a_i$ by $a_i+x_i$ we get a new solution of the first equation in which the sum of the coefficients is $\sum_{i=1}^{n}(a_i+x_i)=M-M=0$.

Let's distinguish two cases:

For $n$ odd there is always the solution $x_i=-1$ for $i=2,...,n$ and $x_1=\frac{1}{2}\sum_{i=2}^{n}p_i$. But then you can add $2M$ to $x_2$ and subtract $3M$ to $x_1$ to get a new solution $(x_1-3M)2+(x_2+2M)3+x_3p_3+...+x_np_n=0$ such that $$(x_1-3M)+(x_2+2M)+x_3+...+x_n=-M$$

For $n$ even we can start from the solution $x_i=-1$ for $i=3,4,...,n$, $x_2=-2$ and $x_1=\frac{1}{2}\sum_{i=2}^{n}x_i$. Then we do the same modification as before. We add $2M$ to $x_2$ and subtract $3M$ from $x_1$.