## 1 + 2 + 3 +…+ m = (m+1) + (m+2) +…+ n

Find all pairs in positive integers   $m, \; n$   such that

$1 \; + \; 2 \; + \; ... \; + \; m \; = \; (m + 1) \; + \; (m + 2) \; + \; ... \; + \; n$

For example,

$1+2 \; = \; 3$

$1+2+3+...+14 \; = \; 105 \; = \; 15+16+17+18+19+20$

$1+2+3+...+84 \; = \; 3570 \; = \; 85+86+87+...+119$

$1+2+3+...+492 \; = \; 121278 \; = \; 493+494+495+...+696$

$1+2+3+...+2870 \; = \; 4119885 \; = \; 2871+2872+2873+...+4059$

$1+2+3+...+16730 = 139954815 = 16731+16732+16733+...+23660$

$1+2+3+...+97512 = 4754343828 = 97513+97514+97515+...+137903$

$1+2+3+...+568344 = 161507735340 = 568345+568346+568347+...+803760$

$1+2+3+...+3312554$
$= 5486508657735 = 3312555+3312556+3312557+...+4684659$

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged , . Bookmark the permalink.

### One Response to 1 + 2 + 3 +…+ m = (m+1) + (m+2) +…+ n

1. paul says:

Here are the next few, though there will be an infinite number of them. This can be reduced to solving m^2 +m = n(n+1)/2 for {m and n}

{m->19306982,n->27304196}
{m->112529340,n->159140519}
{m->655869060,n->927538920}
{m->3822685022,n->5406093003}
{m->22280241074,n->31509019100}
{m->129858761424,n->183648021599}
{m->756872327472,n->1070379110496}
{m->4411375203410,n->6238626641379}
{m->25711378892990,n->36361380737780}
{m->149856898154532,n->211929657785303}
{m->873430010034204,n->1235216565974040}
{m->5090723162050694,n->7199369738058939}
{m->29670908962269962,n->41961001862379596}
{m->172934730611569080,n->244566641436218639}
{m->1007937474707144520,n->1425438846754932240}
{m->5874690117631298042,n->8308066439093374803}
{m->34240203231080643734,n->48422959787805316580}
{m->199566529268852564364,n->282229692287738524679}
{m->1163158972382034742452,n->1644955193938625831496}
{m->6779387305023355890350,n->9587501471344016464299}

Paul.