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Difference between revisions of "1987 OIM Problems/Problem 4"
(solution)
 
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== Problem ==
 
== Problem ==
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We define the succession <math>p_n</math> the following way: <math>p_1=2</math> and for all <math>n</math> more or equal than 2, <math>p_n</math> is the greatest prime divisor of the expression:
+
We define the sequence <math>p_n</math> as follows: <math>p_1=2</math>, and for all <math>n\ge2</math>, <math>p_n</math> is the greatest prime divisor of the expression:
−
<cmath>p_1p_2p_1\cdots p_{n-1} +1</cmath>
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<cmath>p_1p_2p_3\cdots p_{n-1} +1</cmath>
−
Prove that <math>p_n</math> is different than 5
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Prove that <math>p_n\ne5</math>.
  
 
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
 
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
 +
 +
~additional modifications by [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
  
 
== Solution ==
 
== Solution ==
−
{{solution}}
+
We get <math>p_2=3</math> from the definition; thus for all <math>k\ge 3</math>, we must have <math>2,3\nmid p_k</math>. But if we assume for the sake of contradiction that there exists <math>n</math> such that <math>p_n=5</math>, then <math>p_1p_2\cdots p_{n-1}=5^m-1</math> for some positive integer <math>m</math> because <math>5</math> is the minimal prime divisor of <math>p_1p_2\cdots p_{n-1}</math> since <math>2</math> and <math>3</math> are not prime divisors, but <math>5</math> must be the maximal prime divisor by definition. However, taking mod <math>4</math> reveals that the LHS is divisible by <math>2</math> but not <math>4</math> whereas the RHS is divisible by <math>4</math>, a contradiction.
 +
 
 +
~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
  
 
== See also ==
 
== See also ==
 
https://www.oma.org.ar/enunciados/ibe2.htm
 
https://www.oma.org.ar/enunciados/ibe2.htm

Latest revision as of 21:43, 26 August 2025

Problem

We define the sequence $p_n$ as follows: $p_1=2$, and for all $n\ge2$, $p_n$ is the greatest prime divisor of the expression: \[p_1p_2p_3\cdots p_{n-1} +1\] Prove that $p_n\ne5$.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

~additional modifications by eevee9406

Solution

We get $p_2=3$ from the definition; thus for all $k\ge 3$, we must have $2,3\nmid p_k$. But if we assume for the sake of contradiction that there exists $n$ such that $p_n=5$, then $p_1p_2\cdots p_{n-1}=5^m-1$ for some positive integer $m$ because $5$ is the minimal prime divisor of $p_1p_2\cdots p_{n-1}$ since $2$ and $3$ are not prime divisors, but $5$ must be the maximal prime divisor by definition. However, taking mod $4$ reveals that the LHS is divisible by $2$ but not $4$ whereas the RHS is divisible by $4$, a contradiction.

~ eevee9406

See also

https://www.oma.org.ar/enunciados/ibe2.htm

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