Select Page

# Understand the problem

For each positive integer $n$ let $$x_n=p_1+\cdots+p_n$$ where $p_1,\ldots,p_n$ are the first $n$ primes. Prove that for each positive integer $n$, there is an integer $k_n$ such that $x_n
##### Source of the problem
SMO (senior)-2014 stage 2 problem 4

Number Theory
Medium
##### Suggested Book
Excursion in Mathematics

Do you really need a hint? Try it first!

We have $$P_1 = 2 , P_=3 , P_3=5 , P_4 =7 , P_5 = 11 \ and \ so \ on ….$$. Now to understand the expression $x_n , observe . $$For \ n=1 \ , \ 2 < 2^2 < 2+3$$ $$For \ n=2 \ , \ 2+3 < 3^2 < 2+3+5$$ $$For \ n=3 \ , \ 2+3+5 < 4^2 < 2+3 +5+7$$ $$For \ n=4 \ , \ 2+3 +5+7 < 5^2 <2+3 +5+7 +11$$ Now proceed to prove $$\forall n \geq 5$$ .
Observe $$\forall n \geq 5$$ we have $$P_n > (2n-1)$$. [where $$n \in Z^+$$] Then try to use $$x_n = P_1 + P_2 + …+P_5+….. +P_n > 1 +3 + ….+ 9 +… (2n-1) = n^2 \\ \Rightarrow x_n > n^2 , \forall n \geq 5[where \ n \in Z^+]$$ .
Think if $$x_n=P_1 + P_2 + …. + P_5 +…P_n = b^2 for \ some \ n , b \in Z^+$$ , then we are done . If not so , then think $$m$$ be the largest non negative integer such that $$(n+m)^2 < x_n$$ . Now note that the next perfect square is $$(n+m+1)^2$$ . Observe that if we can prove that $$(n+m+1)^2 – (n+m)^2 = (2n+ 2m +1) \geq P_{n+1}$$ , then we are done . Now try to verify this claim .

Suppose our claim is not true i.e. $$P_n < 2n + 2m +1$$ . So, $$P_n < 2n + 2m +1 \\ \Rightarrow 2n+ 2m \geq P_n , \forall n \in Z^+ \\ \Rightarrow (2n +2m-2)+(2n+ 2m -4)+…..2m \geq P_n + P_{n-1}+……+P_1 \\ \Rightarrow n^2 + 2mn -n \geq P_n + P_{n-1}+……+P_1 \\ \Rightarrow n^2 + 2mn -n \geq x_n \\ \Rightarrow n^2 + 2mn +m^2 > n^2 + 2mn -n\geq x_n \\ \Rightarrow (n+m)^2 > x_n$$ . Contradiction! since we have assumed $$x_n = P_1 + P_2+……+P_{n-1} > (n+m)^2$$ . Thus ,$$(n+m+1)^2 \in (x_n , x_{n+1})$$ .

# Connected Program at Cheenta

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

# Similar Problems

## RMO 2019 (Maharashtra Goa) Adding GCDs

Can you add GCDs? This problem from RMO 2019 (Maharashtra region) has a beautiful solution. We also give some bonus questions for you to try.

## Number Theory, Ireland MO 2018, Problem 9

This problem in number theory is an elegant applications of the ideas of quadratic and cubic residues of a number. Try with our sequential hints.

## Number Theory, France IMO TST 2012, Problem 3

This problem is an advanced number theory problem using the ideas of lifting the exponents. Try with our sequential hints.

## Algebra, Austria MO 2016, Problem 4

This algebra problem is an elegant application of culminating the ideas of polynomials to give a simple proof of an inequality. Try with our sequential hints.

## Number Theory, Cyprus IMO TST 2018, Problem 1

This problem is a beautiful and simple application of the ideas of inequality and bounds in number theory. Try with our sequential hints.

## Number Theory, South Africa 2019, Problem 6

This problem in number theory is an elegant applciations of the modulo technique used in the diophantine equations. Try with our sequential hints

## Number Theory, Korea Junior MO 2015, Problem 7

This problem in number theory is an elegant application of the ideas of the proof of infinitude of primes from Korea. Try with our sequential hints.

## Inequality, Israel MO 2018, Problem 3

This problem is a basic application of triangle inequality along with getting to manipulate the modulus function efficently. Try with our sequential hints.

## Number Theory, Greece MO 2019, Problem 3

This problem is a beautiful application of prime factorization theorem, and reveal how important it is. Try with our sequential hints.

## Algebra, Germany MO 2019, Problem 6

This problem is a beautiful application of algebraic manipulations, ideas of symmetry, and vieta’s formula in polynomials. Try with our sequential hints.