Product of first m primes

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Product of first m primes

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Source: European Mathematical Cup 2013, Junior Division, P1

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#1 h Jul 3, 2014, 7:57 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

For $m\in \mathbb{N}$ define $m?$ be the product of first $m$ primes. Determine if there exists positive integers $m,n$ with the following property :
$m?=n(n+1)(n+2)(n+3)$

Proposed by Matko Ljulj

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#2 h Jul 3, 2014, 8:54 AM • 2 Y

Y by Adventure10, Mango247

Since RHS is divisible by $4$ whereas LHS is divisible by $2$ only we see no such pair exists. Nice and short problem, I must say.
Cheers, grammarphobic.

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#3 h Jul 3, 2014, 10:24 PM • 2 Y

Y by Adventure10, Mango247

Unexpectedly, the problem was easy. But I guess it can be transformed into a nice one.

For example, we can try $2^2m?=n(n+1)(n+2)(n+3)$ . Because, the case $2m?$ is trivial for a similar reason: right side would be divisible by $8$ . Since there are two consecutive even integers.

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#4 h Apr 26, 2021, 12:41 PM

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joybangla wrote:

For $m\in \mathbb{N}$ define $m?$ be the product of first $m$ primes. Determine if there exists positive integers $m,n$ with the following property :
$m?=n(n+1)(n+2)(n+3)$
Proposed by Matko Ljulj

By contradiction, assume that exists an $m$ and an $n$ such that works:
Since they are both positive integers $n(n+1)(n+2)(n+3) \equiv 0 \pmod 4$ .
Note that the only pair number and prime is $2$ and then $m?=2 \cdot 3 \cdots p$ where $p$ is the $m$ order prime and this means $m? \equiv 2 \pmod 4$ .
Hence $2 \equiv 0 \pmod 4 \; \text{contradiction!!}$ .
That means no such integers, then we are done :blush:

:blush:

This post has been edited 1 time. Last edited by MathLuis, Apr 26, 2021, 12:41 PM

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#5 h Apr 4, 2025, 10:48 AM

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When we have $4$ consecutive numbers exactly one of them is divisible by $4$ , or said differently exactly one of them is $0(mod 4)$ , so the $RHS$ is $0 (mod 4)$ , but the $LHS$ is clearly $2(mod4)$ .
$Q.E.D$

(It is very well-known that there is only one even prime ( $2$ )).

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#6 h Apr 4, 2025, 11:33 AM

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Take mod 4 on both sides. RHS gives 0 $(4 | n+i$ for some $0 <= i <= 3 )$ .
LHS gives 2( for $m >= 1$ , product is a mulitple of 2).
So, no such $m,n$ can exist.

This post has been edited 1 time. Last edited by Erratum, Apr 4, 2025, 11:34 AM

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#7 h Apr 4, 2025, 12:03 PM

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The LHS is not divisible by $4$ , but the RHS is $4 \binom{n+3}{4}$ , which is divisible by $4$ , so such positive integers $m,n$ don't exist.

This post has been edited 2 times. Last edited by megarnie, Apr 4, 2025, 12:04 PM

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