Triangular number & sum of four squares in 2 ways

 
 

An integer of the form   T_{n} \; = \; n \, (n + 1)/2   is called a Triangular number.

 

1 + 4 + 6 + 7 \; = \; 2 + 3 + 5 + 8 \; = \; 18

1^2 + 4^2 + 6^2 + 7^2 \; = \; 2^2 + 3^2 + 5^2 + 8^2 \; = \; 102
 

substituting   (1, 4, 6, 7)   and   (2, 3, 5, 8)   for a triangular number :

T_1 = 1   …..   T_4 = 10   …..   T_6 = 21   …..   T_7 = 28
T_2 = 3   …..   T_3 = 6   ……   T_5 = 15   …..   T_8 = 36

1 + 10 + 21 + 28 \; = \; 3 + 6 + 15 + 36 \; = \; 60

Similarly,

 

5 + 8 + 10 + 11 \; = \; 6 + 7 + 9 + 12 \; = \; 34

5^2 + 8^2 + 10^2 + 11^2 \; = \; 6^2 + 7^2 + 9^2 + 12^2 \; = \; 310

T_5 = 15   …..   T_8 = 36   …..   T_{10} = 55   …..   T_{11} = 66
T_6 = 21   …..   T_7 = 28   …..   T_9 = 45   ……..   T_{12} = 78

15 + 36 + 55 + 66 \; = \; 21 + 28 + 45 + 78 \; = \; 172

 

By adding

1 + 4 + \; 6 \; + \; 7 \; = \; 2 + 3 + 5 + 8
5 + 8 + 10 + 11 \; = \; 6 + 7 + 9 + 12
———————————————————-
6 + 12 + 16 + 18 \; = \; 8 + 10 + 14 + 20 \; = \; 52

6^2 + 12^2 + 16^2 + 18^2 \; = \; 8^2 + 10^2 + 14^2 + 20^2 \; = \; 760

T_6 = 21   …..   T_{12} = 78   …..   T_{16} = 136 …..   T_{18} = 171

T_8 = 36   …..   T_{10} = 55   …..   T_{14} = 105   …..   T_{20} = 210
 

21 + 78 + 136 + 171 \; = \; 36 + 55 + 105 + 210 \; = \; 406

 
Find other examples.
 

Paul found

TRIANGULAR 1

1+4+6+7 \; = \; 2+3+5+8 \; = \; 18
1+4+7+8 \; = \; 2+3+6+9 \; = \; 20
2+8+13+15 \; = \; 4+6+11+17 \; = \; 38
2+5+7+8 \; = \; 3+4+6+9 \; = \; 22
3+9+13+15 \; = \; 5+7+11+17 \; = \; 40
1+6+7+8 \; = \; 3+4+5+10 \; = \; 22
2+10+13+15 \; = \; 5+7+10+18 \; = \; 40
2+9+10+11 \; = \; 5+6+7+14 \; = \; 32
5+8+10+11 \; = \; 6+7+9+12 \; = \; 34
7+17+20+22 \; = \; 11+13+16+26 \; = \; 66
1+7+8+14 \; = \; 2+4+11+13 \; = \; 30
6+15+18+25 \; = \; 8+11+20+25 \; = \; 64
3+6+11+12 \; = \; 4+5+10+13 \; = \; 32
8+14+21+23 \; = \; 10+12+19+25 \; = \; 66

1^2+4^2+6^2+7^2 = 2^2+3^2+5^2+8^2 = 102
1^2+4^2+7^2+8^2 = 2^2+3^2+6^2+9^2 = 130
2^2+8^2+13^2+15^2 = 4^2+6^2+11^2+17^2 = 462
2^2+5^2+7^2+8^2 = 3^2+4^2+6^2+9^2 = 142
3^2+9^2+13^2+15^2 = 5^2+7^2+11^2+17^2 = 484
1^2+6^2+7^2+8^2 = 3^2+4^2+5^2+10^2 = 150
2^2+10^2+13^2+15^2 = 5^2+7^2+10^2+18^2 = 498
2^2+9^2+10^2+11^2 = 5^2+6^2+7^2+14^2 = 306
5^2+8^2+10^2+11^2 = 6^2+7^2+9^2+12^2 = 310
7^2+17^2+20^2+22^2 = 11^2+13^2+16^2+26^2 = 1222
1^2+7^2+8^2+14^2 = 2^2+4^2+11^2+13^2 = 310
6^2+15^2+18^2+25^2 = 8^2+11^2+20^2+25^2 = 1210
3^2+6^2+11^2+12^2 = 4^2+5^2+10^2+13^2 = 310
8^2+14^2+21^2+23^2 = 10^2+12^2+19^2+25^2 = 1230

  T_1 = 1   ……   T_2 = 3   ……   T_3 = 6
  T_4 = 10   …..   T_5 = 15   …..   T_6 = 21
  T_7 = 28   …..   T_8 = 36   …..   T_9 = 45

T_{10} = 55   ……   T_{11} = 66   ……   T_{12} = 78
T_{13} = 91   ……   T_{14} = 105   …..   T_{15} = 120
T_{16} = 136   …..   T_{17} = 153   …..   T_{18} = 171
T_{19} = 190   …..   T_{20} = 210   …..   T_{21} = 231
T_{22} = 253   …..   T_{23} = 276   …..   T_{24} = 300
T_{25} = 325   …..   T_{26} = 351   …..   T_{27} = 378
T_{28} = 406   …..   T_{29} = 435   …..   T_{30} = 465

(1,4,6,7)   ——->   (1, 10, 21, 28)
{2,3,5,8)   ——->   (3, 6, 15, 36)

1 + 10 + 21 + 28   =   3 + 6 + 15 + 36   =   60

(1,4,7,8)   ——->   (1, 10, 28, 36)
(2,3,6,9)   ——->   (3, 6, 21, 45)

1 + 10 + 28 + 36 = 3 + 6 + 21 + 45   =   75

(2,8,13,15)   —–>   (3, 36, 91, 120)
(4,6,11,17)   —–>   (10, 21, 66, 153)

3 + 36 + 91 + 120   =   10 + 21 + 66 + 153   =   250

(2, 5, 7, 8)   ———->   (3, 15, 28, 36)
(3 ,4 ,6, 9)   ———->   (6, 10, 21, 45)

3 + 15 + 28 + 36   =   6 + 10 + 21 + 45   =   82

(3, 9, 13, 15)   ——–>   (6, 45, 91, 120)
(5, 7, 11, 17)   ——–>   (15, 28, 66, 153)

6 + 45 + 91 + 120   =   15 + 28 + 66 + 153   =   262

(1, 4, 6, 7)   ———->   (1, 10, 21, 28)
(2, 3, 5, 8)   ———->   (3, 6, 15, 36)

1 + 10 + 21 + 28   =   3 + 6 + 15 + 36   =   60

(1, 6 ,7, 8)   ———->   (1, 21, 28, 36)
(3, 4, 5, 10)   ———>   (6, 10, 15, 55)

1 + 21 + 28 + 36   =   6 + 10 + 15 + 55   =   86

(2, 10, 13, 15)   ——->   (3, 55, 91, 120)
(5, 7, 10, 18)   ——–>   (15, 28, 55, 171)

3 + 55 + 91 + 120   =   15 + 28 + 55 + 171   =   269

 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

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

3 Responses to Triangular number & sum of four squares in 2 ways

  1. Paul says:

    There are 1114 solutions with the sum of squares <= 310, Here are a few.

    The isolated one is the example given above so you get the format, they all pass each test for equal sum, equal sum of squares and T Numbers etc

    {{{1,4,6,7},{2,3,5,8}},{{1,4,7,8},{2,3,6,9}}} , {{2,8,13,15},{4,6,11,17}}
    {{{1,4,6,7},{2,3,5,8}},{{2,5,7,8},{3,4,6,9}}} , {{3,9,13,15},{5,7,11,17}}
    {{{1,4,6,7},{2,3,5,8}},{{1,6,7,8},{3,4,5,10}}} , {{2,10,13,15},{5,7,10,18}}

    {{{1,4,6,7},{2,3,5,8}},{{5,8,10,11},{6,7,9,12}}} , {{6,12,16,18},{8,10,14,20}}

    {{{2,9,10,11},{5,6,7,14}},{{5,8,10,11},{6,7,9,12}}} , {{7,17,20,22},{11,13,16,26}}
    {{{1,7,8,14},{2,4,11,13}},{{5,8,10,11},{6,7,9,12}}} , {{6,15,18,25},{8,11,20,25}}
    {{{3,6,11,12},{4,5,10,13}},{{5,8,10,11},{6,7,9,12}}} , {{8,14,21,23},{10,12,19,25}}

    Paul.

    • benvitalis says:

      Thanks for the drop box folder

    • Paul says:

      Here is the MMA code, it is set between 100 and 150 for the sum of squares.

      Clear[a, b, c, d]; lt = {}; cnt = 0; lts = {}; Do[
       p = Solve[
         a^2 + b^2 + c^2 + d^2 == n && a < b < c < d && a > 0, {a, b, c, d},
          Integers]; p = {a, b, c, d} /. p; 
       f = Cases[
         e = Subsets[p, {2}], {{a_, b_, c_, d_}, {e_, f_, g_, h_}} /; 
          a + b + c + d == 
            e + f + g + h && (a (a + 1))/2 + (b (b + 1))/2 + (c (c + 1))/
             2 + (d (d + 1))/2 == (e (e + 1))/2 + (f (f + 1))/2 + (
             g (g + 1))/2 + (h (h + 1))/2 && 
           Length[Union[{a, b, c, d, e, f, g, h}]] == 8]; 
       If[Length[f] > 0, f = Partition[Partition[Flatten[f], 4], 2]; 
        AppendTo[lt, f]], {n, 100, 150}];
      lt = Partition[Partition[Flatten[lt], 4], 2]; Do[
       AppendTo[lts, {lt[[q]], lt[[w]]}], {q, 1, Length[lt] - 1}, {w, q + 1,
         Length[lt]}]; Do[j = lts[[q, 1]] + lts[[q, 2]]; 
       If[Length[Union[Total /@ (lts[[q, 1]] + lts[[q, 2]])]] == 1 && 
         Length[Union[Total /@ ((lts[[q, 1]] + lts[[q, 2]])^2)]] == 
          1 && (j[[1, 1]] (j[[1, 1]] + 1))/2 + (j[[1, 2]] (j[[1, 2]] + 1))/
           2 + (j[[1, 3]] (j[[1, 3]] + 1))/2 + (j[[1, 4]] (j[[1, 4]] + 1))/
           2 == (j[[2, 1]] (j[[2, 1]] + 1))/2 + (j[[2, 2]] (j[[2, 2]] + 1))/
           2 + (j[[2, 3]] (j[[2, 3]] + 1))/2 + (j[[2, 4]] (j[[2, 4]] + 1))/
           2, Print[lts[[q]], " , ", j]; cnt++], {q, 1, 
        Length[lts]}]; Print[cnt]

      Paul.

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