# Existence of perfect square between the sum of the first $n$ and $n + 1$ prime numbers

Let $A_n$ be the sum of the first $n$ prime numbers. Prove that there is a perfect square between $A_n$ and $A_{n+1}$.

This is how I prove the conjucture:

Using Proof by contradiction, suppose there is no perfect square between A_n and A_(n+1).

It implies that k^2 < A_n and (〖k+1)〗^2>A_(n+1) for any natural number k.

Let A_n=p_1+p_2+p_3+⋯+p_n A_(n+1)=p_1+p_2+p_3+⋯+p_(n+1)

The difference between (〖k+1)〗^2and k^2 is 2k+1 and the difference between A_n and A_(n+1)is p_(n+1), thus p_(n+1)<2k+1.

Since k^2= 1+3+5+...+(2k-1), thus p_n ≥ 2k-1 and p_(n+1) ≥ 2k+1. It then contradicts p_(n+1)<2k+1.

Therefore there is a perfect square between A_n and A_(n+1)where A_n is the sum of the first n prime numbers.

• including them? – evil999man Apr 21 '14 at 16:47
• @user144691, just FYI you should generally wait at least a day before accepting an answer. – 6005 Apr 21 '14 at 17:05
• What is the source of the problem? What other material is discussed in the book before this exercise? – Will Jagy Apr 21 '14 at 18:04
• Judging from this other post, this OP is asking questions without knowing whether they are true or false, although worded as though a proof is available and easy: math.stackexchange.com/questions/763147/… Voting to close – Will Jagy Apr 21 '14 at 18:10
• What is your argument for $p_n \geqslant 2k-1$? – Daniel Fischer Jul 20 '14 at 10:57

## 1 Answer

This statement appears to be true, and could perhaps be proved with fairly basic analytic estimates, but it is by no means an elementary question. Here is data up to the point where the sum of consecutive primes exceeds 10,000. Note that, when the sum is exactly 100, the square 81 could have been chosen instead. Posted with CW status as I voted to close the question.

  April 21
1
2         2
4
3         5
9
5        10
16
7        17
25
11        28
36
13        41
49
17        58
64
19        77
100
23       100  =-=-=-=-=
121
29       129
144
31       160
196
37       197
225
41       238
256
43       281
324
47       328
361
53       381
400
59       440
484
61       501
529
67       568
625
71       639
676
73       712
784
79       791
841
83       874
961
89       963
1024
97      1060
1156
101      1161
1225
103      1264
1369
107      1371
1444
109      1480
1521
113      1593
1681
127      1720
1849
131      1851
1936
137      1988
2116
139      2127
2209
149      2276
2401
151      2427
2500
157      2584
2704
163      2747
2809
167      2914
3025
173      3087
3249
179      3266
3364
181      3447
3600
191      3638
3721
193      3831
3969
197      4028
4225
199      4227
4356
211      4438
4624
223      4661
4761
227      4888
5041
229      5117
5329
233      5350
5476
239      5589
5776
241      5830
5929
251      6081
6241
257      6338
6561
263      6601
6724
269      6870
7056
271      7141
7396
277      7418
7569
281      7699
7921
283      7982
8100
293      8275
8464
307      8582
8836
311      8893
9025
313      9206
9409
317      9523
9801
331      9854
10000
337     10191


The thing would appear to follow from effective bounds on the size of the $n$-th prime in Rosser and Schoenfeld (1962) available online. The resulting inequality confirms the statement for a specific lower bound $n \geq N,$ and a computer run confirms it below $n.$ I expect i will figure out $N$ tomorrow, it is not immediate...