If $p^2+q^2+7pq = r^2$ ($r$ being any integer), then $(p+q)^2 + 5pq = r^2$. So $5pq = r^2-(p+q)^2 = (r+p+q)(r-p-q)$.
Since $p, q$ and $5$ are all prime, it follows that one of the factors on the right-hand side is equal to one of them, and the other factor is the product of the other two. As clearly visible, $r+p+q$ is greater than any of the numbers $p, q$ and $5$. So it must be the product of two of those numbers (maybe three too), and the other factor $r-p-q$ must be equal to $p, q$ or $5$ (or $1$).
Now, different cases arise:
CASE $1$:
If $r-p-q = p$ then $r=2p+q$, and thus original equation becomes $p^2+q^2+7pq = (2p+q)^2$, which simplifies to $p=q$, Same if $r-p-q = q$.
CASE $2$:
If $r-p-q = 5$ then $r = p+q+5$, and the equation $5pq = (r+p+q)(r-p-q)$ becomes $pq = 2(p+q)+5$. You can write this as $(p-2)(q-2)=9$, and the only solutions to this are $p=q=5$ and $p=3;q=11$ (or the other way round).
CASE $3$: $r+p+q$ might be equal to the product of all three numbers $5pq$, with $r-p-q = 1$. But then the equation becomes $2p+2q+1 = 5pq$, which is clearly impossible because the right-hand side is visibly greater than the left-hand side (Though there are solutions like $(1,1)$, but $1$ is not prime ).
So, to summarise the whole answer, We can say that only solutions are:
$(p,q)=(p,p),(3,11),(11,3)$
SOURCE
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...
Best Answer
looking at prime factorisation of 2013 the problem becomes; $$N(N+3*11*61) = m^2$$
if we let $N$ equal the product of two of the primes and anyother number say $n$ we have $$P_1*P_2*n(P_1*P_2*(P_3 + n) = m^2$$ where $p_i$ is one of ${3,11,61}$
It is easy to see that if we take $N = 11*61*1$ we will have $$11*61*11*61*(3+1)$$ which gives $$(11*61*2)^2$$ so $m=11*61*2 = 1342$ and $N=671$