Prime numbers 127, 3697, 5227

$127, \; 3697, \; 5227$   are prime numbers such that the numbers

$p \,(p+1)$,    $q \,(q+1)$,    $r \,(r+1)$

form an arithmetic progression :

$3697(3697 + 1) \; - \; 127(127 + 1)$
$= \; 5227(5227 + 1) \; - \; 3697(3697 + 1)$
$= \; 13655250$

Find other such numbers

The problem may be expressed as follows: find three triangular numbers with prime indices, which form an increasing arithmetic progression.

Note that

1783,   3697,   3001   are primes congruent to 14 mod 29, that is,   $29 \, n + 14$,   and
2521,   5227,   4243   are primes congruent to 20 mod 41,   $41 \, n + 20$

$n = 61$,    $1783 = 29\times 61 + 14$,    $2521 = 41\times 61 + 20$

$1783 (1783 + 1) - 61 (61 + 1) = 2521 (2521 + 1) - 1783 (1783 + 1) = 3177090$

$n = 127$,    $3697 = 29\times 127 + 14$,    $5227 = 41\times 127 + 20$

$3697 (3697 + 1) - 127 (127 + 1) = 5227(5227 + 1) - 3697(3697 + 1) = 13655250$

$n = 103$,    $3001 = 29\times 103 + 14$,    $4243 = 41\times 103 + 20$

$3001 (3001 + 1) - 103 (103 + 1) = 4243 (4243 + 1) - 3001 (3001 + 1) = 8998290$

In these solutions, the numbers   $n$,   $29 \,n+14$,   and   $41 \,n+20$   are all primes

and, the numbers   $n$,   $29 \,n+14$, and   $41 \,n+20$   form an arithmetic progression:

$(29 \, n+14) \,(29 \, n+15) \; - \; n \,(n+1) \; = \; 210 \, (2 \, n + 1)^2$
$(41 \, n+20) \,(41 \,n+21) \; - \; (29 \, n+14) \,(29 \, n+15) \; = \; 210 \, (2 \, n + 1)^2$

$(29 \, n+14) \,(29 \, n+15) \; - \; n \,(n+1)$
$= \; (41 \, n+20)(41 \, n+21) \; - \; (29 \, n+14)(29 \, n+15)$

There are other solutions that Paul have found

math grad - Interest: Number theory
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3 Responses to Prime numbers 127, 3697, 5227

1. paul says:

Here are a few more

757(757 + 1) – 61(61 + 1) = 1069(1069 + 1) – 757(757 + 1) = 570024
1783(1783 + 1) – 61(61 + 1) = 2521(2521 + 1) – 1783(1783 + 1) = 3177090
1657(1657 + 1) – 109(109 + 1) = 2341(2341 + 1) – 1657(1657 + 1) = 2735316
397(397 + 1) – 127(127 + 1) = 547(547 + 1) – 397(397 + 1) = 141750
1033(1033 + 1) – 331(331 + 1) = 1423(1423 + 1) – 1033(1033 + 1) = 958230
1447(1447 + 1) – 1249(1249 + 1) = 1621(1621 + 1) – 1447(1447 + 1) = 534006
2377(2377 + 1) – 1249(1249 + 1) = 3121(3121 + 1) – 2377(2377 + 1) = 4091256
2677(2677 + 1) – 1291(1291 + 1) = 3559(3559 + 1) – 2677(2677 + 1) = 5501034

Paul.

2. paul says:

and a few more

3001(3001 + 1) – 103(103 + 1) = 4243(4243 + 1) – 3001(3001 + 1) = 8998290
3697(3697 + 1) – 127(127 + 1) = 5227(5227 + 1) – 3697(3697 + 1) = 13655250
4813(4813 + 1) – 433(433 + 1) = 6793(6793 + 1) – 4813(4813 + 1) = 22981860
4447(4447 + 1) – 727(727 + 1) = 6247(6247 + 1) – 4447(4447 + 1) = 19251000
3253(3253 + 1) – 1201(1201 + 1) = 4441(4441 + 1) – 3253(3253 + 1) = 9141660
4831(4831 + 1) – 1933(1933 + 1) = 6553(6553 + 1) – 4831(4831 + 1) = 19604970
3373(3373 + 1) – 2203(2203 + 1) = 4231(4231 + 1) – 3373(3373 + 1) = 6525090
4051(4051 + 1) – 2221(2221 + 1) = 5281(5281 + 1) – 4051(4051 + 1) = 11479590
4363(4363 + 1) – 2473(2473 + 1) = 5653(5653 + 1) – 4363(4363 + 1) = 12921930
5227(5227 + 1) – 2521(2521 + 1) = 6949(6949 + 1) – 5227(5227 + 1) = 20968794
3607(3607 + 1) – 3301(3301 + 1) = 3889(3889 + 1) – 3607(3607 + 1) = 2114154
5227(5227 + 1) – 4861(4861 + 1) = 5569(5569 + 1) – 5227(5227 + 1) = 3692574

Paul.