Sum of consecutive primes

Algebra Level 3

For each positive integer $n$ , $S_n$ is the sum of the first $n$ consecutive prime numbers.

For example:

$\begin{aligned} n=1 & \implies S_1= 2 \\ n=2 & \implies S_2=2+3 = 5\end{aligned}$

Arrange these numbers in a sequence of ascending order: $S_1, S_2, S_3,...$ . Is the statement below true or false?

In the sequence above, we can always find two consecutive numbers which are both perfect squares.

False True

1 solution

Adrian Klaeger
Aug 2, 2020

By contradiction assume there exist integers $\sqrt{S_{n-1}}$ and $\sqrt{S_n}$ .

By definition $S_n - S_{n-1} = p_n$ , where $p_n$ is the nth prime.

$(\sqrt{S_n} + \sqrt{S_{n-1}})(\sqrt{S_n} - \sqrt{S_{n-1}}) = p_n$

Since $p_n$ is prime, the two factors above must be:

$(\sqrt{S_n} + \sqrt{S_{n-1}}) = p_n$

$(\sqrt{S_n} - \sqrt{S_{n-1}}) = 1$

Adding both equations gives:

$2 \sqrt{S_n} = p_n + 1$

$S_n = (\frac{p_n + 1}{2})^2$

The expression $(\frac{p_n + 1}{2})^2$ can be interpreted as the sum of all odd numbers up to $p_n$ .

However, these values will never be equal, since $S_n$ will be greater than the sum of odd numbers up to $n = 4$ due to the even prime $2$ , and less afterwards due to non-prime odd numbers being skipped, starting with $9$ .

Excuse me, but I don't understand why $(\frac{p_n+1}{2})^2$ can be interpreted as sum of all odd numbers up to $p_n$ ?

Tin Le - 10 months, 2 weeks ago

If $o_n = 2n - 1$ is the nth odd number, then $n = \frac{o_n + 1}{2}$ .

The sum of the first $n$ odd numbers is $n^2$ (see properties of square numbers or sum of arithmetic progression ).

Therefore, the sum of all odd numbers up to $o_n$ is $(\frac{o_n + 1}{2})^2$

Adrian Klaeger - 10 months, 1 week ago

Sum of consecutive primes

1 solution

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