## To make {(3A + 1), (7A + 1)} squares

$3 \, A \; + \; 1 \; = \; B^2$
$7 \, A \; + \; 1 \; = \; C^2$

$A$   is positive integer

Here are the first few solutions:

$3(120) + 1 = 19^2$
$7(120) + 1 = 29^2$

$3(2760) + 1 = 91^2$
$7(2760) + 1 = 139^2$

$3(1454640) + 1 = 2089^2$
$7(1454640) + 1 = 3191^2$

$3(33393360) + 1 = 10009^2$
$7(33393360) + 1 = 15289^2$

$3(17598237480) + 1 = 229771^2$
$7(17598237480) + 1 = 350981^2$

$3(403992869400) + 1 = 1100899^2$
$7(403992869400) + 1 = 1681651^2$

$3(212903475581280) + 1 = 25272721^2$
$7(212903475581280) + 1 = 38604719^2$

$3(4887505700610720) + 1 = 121088881^2$
$7(4887505700610720) + 1 = 184966321^2$

$3(2575706229984090840) + 1 = 2779769539^2$
$7(2575706229984090840) + 1 = 4246168109^2$

$3(59129043561995624040) + 1 = 13318676011^2$
$7(59129043561995624040) + 1 = 20344613659^2$

$3(31160893757444055403920) + 1 = 305749376569^2$
$7(31160893757444055403920) + 1 = 467039887271^2$

$3(715343164125517359028080) + 1 = 1464933272329^2$
$7(715343164125517359028080) + 1 = 2237722536169^2$

$3(376984490101851952292536200) + 1 = 33629651653051^2$
$7(376984490101851952292536200) + 1 = 51370141431701^2$

$3(8654221540461465447526090680) + 1 = 161129341280179^2$
$7(8654221540461465447526090680) + 1 = 246129134364931^2$

$3(4560758330091311161391047546560) + 1 = 3698955932459041^2$
$7(4560758330091311161391047546560) + 1 = 5650248517599839^2$

$3(104698771481159644858653286021440) + 1 = 17722762607547361^2$
$7(104698771481159644858653286021440) + 1 = 27071967057606241^2$

$3(55176053900460192328656940925749560) + 1 = 406851522918841459^2$
$7(55176053900460192328656940925749560) + 1 = 621475966794550589^2$

$3(1266645728724847843038522006761293320) + 1 = 1949342757488929531^2$
$7(1266645728724847843038522006761293320) + 1 = 2977670247202321579^2$