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Here is a conjecture of mine. Prove that for any positive integer $k$, there exists a prime $p$ such that $${x \choose k}\equiv -1\pmod{p}$$ has an integer solution.

When $k$ is odd, we can take $x$ to be $p-1$, the question can be easily solved because $${p-1 \choose k}\equiv (-1)^k\equiv -1\pmod{p}.$$ However, when $k$ is even, I cannot find any pattern. Here are results for $k=4$ and $k=6$.

$$x=30\implies {x \choose 4}\equiv -1\pmod{193},$$ $$x=34\implies {x \choose 6}\equiv -1\pmod{97}.$$

Edit: See Peter's comment. It is an obvious result. Still, I am very curious about if $p$ can be arbitrarily large. If we use Peter's hint, we need to prove that given $k$, the prime factor of the sequence $$\left\{{x\choose k}+1\right\}_{x=1}^{\infty}$$can be arbitrarily large. I feel like that the proof is similar to Primes dividing a polynomial.

MathEric
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    This is obvious : just choose a prime factor of $k+2$. Then, $x=k+1$ and this prime number do the job. In fact, $x$ can be an arbitary integer greater than $k$ , $p$ only needs to be a prime factor of $\binom{x}{k}+1$ – Peter Jul 01 '20 at 12:15
  • You should see my comments on the accepted answer on this question https://math.stackexchange.com/q/3728157/675453 (sorry, I can't quote my two comments, because my browser keeps crashing on that page) hope that helps you – UmbQbify Jul 01 '20 at 12:20
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    @Peter Sorry, what I really want to ask is that if there exist infinitely many such primes, but your solution seems works for that, am I correct? – MathEric Jul 01 '20 at 12:22
  • A little work has still to be done, but it should be not too difficult to finish the proof. – Peter Jul 01 '20 at 12:26
  • This is in some ways totally natural, the polynomial $f(x)=\binom xk + 1$ giving an integer number $f(n)=\prod p_i^{n_i}$ so the property is valid for all $p_i$. – Piquito Jul 01 '20 at 18:38

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The question you link answers it.

Fix $k$ and consider the polynomial $f(x)={x\choose k}+1$. This is a nonconstant polynomial with rational coefficients. Indeed $$ f(x)=\frac{x(x-1)\ldots(x-k+1)}{k!}+1 $$ If we set $g(x)=k!f(x)$ then $g(x)$ has integer coefficients.

By the question you link (Primes dividing a polynomial) there is an infinite set $P$ of primes such that for any $p\in P$ there is some integer $x_p$ such that $p$ divides $g(x_p)$. Recall that $g(x_p)=k!f(x_p)$.

Let $Q$ be the set of primes $p\in P$ that do not divide $k!$. Then $Q$ is still an infinite set and if $p\in Q$ then $p$ divides $f(x_p)$.

So for infinitely many primes (those in $Q$) there is a solution to ${x\choose k}\equiv -1\space (\textrm{mod } p)$

halrankard
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