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I have studied the Lucas theorem and I encountered the following facts.

How to deduce the following facts from The Lucas theorem?

(1) If d, q > 1 are integers such that , $$\binom{nd}{md}$$ $\equiv$ $\binom{n}{m}$ (mod p) for every pair of integers n greater than or equal to m and m greater than or equal to 0, then d and q are powers of the same prime p.

(2) Let n and r be non-negative integers and p is greater than or equal to 5 be a prime, then $$\binom{np}{rp}$$ $\equiv$ $\binom{n}{r}$ (mod $p^3$). Where we set $$\binom{n}{r}$$ = 0, if n < r.

(3) Let N, R, n and r be non-negative integers and p is greater than or equal to 5 be a prime. Let n, r < p, then $$\binom{Np^3+n}{Rp^3 + r}$$ $\equiv$ $\binom{N}{R}$ $\binom{n}{r}$(mod $p^3$).

(4) If p is a prime and x is a positive integer and $p^x$ divides $$\left\lfloor\frac{n}{p}\right\rfloor\, $$ then, $p^x$ also divides $$\binom{n}{p}$$

I observed the above facts during my study on LUCAS Theorem. If any one can justify the all above four consequence problems, I am very glad and thankful to them. A waiting a good response on these interesting questions. Once gain thank you for this opportunity to post my questions.

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    $\begingroup$ There is no q in the equation of (1). $\endgroup$ – Karolis Juodelė Aug 6 '12 at 11:11
  • $\begingroup$ I'll delete my wrong answer so as not to make this problem look solved. The link to the paper - dms.umontreal.ca/~andrew/PDF/BinCoeff.pdf. The answer to (1) - "I'll assume $\binom{nd}{md}\equiv \binom{nq}{mq}$. As multiplication by a power of $p$ only shifts digits in base $p$, using Lucas theorem both sides expand the same way, except for some terms $\binom{0}{0}$." $\endgroup$ – Karolis Juodelė Aug 9 '12 at 13:53
  • $\begingroup$ @Karolis Juodele, you are very much correct as there is no term q in the first question. Please leave the first question and see the others. I am not able to go further, even by looking your suggested PDF. Plz, try to answer the rest. Once again thank you so much. $\endgroup$ – BMSA Aug 11 '12 at 13:30
  • $\begingroup$ @Karolis Juodele! I am looking this link every time. But, no one is looking this questions. please look and find the way to complete these questions. $\endgroup$ – BMSA Aug 15 '12 at 11:45
  • $\begingroup$ I'll say this. Your statements are not direct consequences of Lucas theorem. They are fairly complex. I am unlikely to solve them. You may be lucky enough to find them in some literature. If not, consider starting a bounty. $\endgroup$ – Karolis Juodelė Aug 15 '12 at 18:39

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