1.Introduction

Andrew Bremner and Allan Macleod[1] showed the positive integer solution of a/(b+c) +b/(c+a) +c(a+b)=n.
We treat diophantine equation a/(b+c+d) +b/(a+c+d) +c/(a+b+d)+ d/(a+b+c) = n, 
and give a condition for positive integer solution of a/(b+c+d) +b/(a+c+d) +c/(a+b+d)+ d/(a+b+c) = 2.
It seems that there are infinitely many positive integer solutions for n=2.
Furthermore, we show the numeric solutions for n<=30.
As a result of searching the solutions, solutions for n=8,27 were not found.


2.Theorem

Diophantine equation a/(b+c+d) +b/(a+c+d) +c/(a+b+d)+ d/(a+b+c) = 2 has positive integer solutions,
if v^2 = 361b^4-38b^3-11099b^2+16500b+22500 has the rational solutions with 0 < b <1 or 3 < b <75/19.

Proof.

a/(b+c+d) +b/(a+c+d) +c/(a+b+d)+ d/(a+b+c) = 2..........................................................................(1)

If there are rational solutions with a,b,c,d>0, then there are positive integer solutions of equation (1).

Let s=a+b+c+d.

(ns^2-ncs+4cb+nbc-3sb-nbs+2s^2-3sc)a^2
+(-nb^2s+2ncs^2+2nbs^2+ncb^2+nc^2b-nc^2s+4cb^2-10scb-3sc^2-3ncsb+5s^2b-2s^3+4c^2b-3sb^2+5s^2c-ns^3)a
-2bs^3+2s^2c^2-2cs^3-3scb^2+2ncs^2b-ncs^3-ncsb^2+nb^2s^2-nc^2sb+nc^2s^2+s^4-3bsc^2+2s^2b^2-nbs^3+5bs^2c=0...............(2)

Discriminant for a must be perfect square number, we obtain

v^2 = (8nc^2+n^2s^2-14nsc+16c^2-24sc+n^2c^2+9s^2-2n^2cs+6ns^2)b^4
    +(-6s^3-30nsc^2+16nc^3+2n^2c^3-4n^2c^2s-56sc^2+2n^2s^2c+32c^3+32s^2c-2ns^3+16ns^2c)b^3
    +(-30nc^3s-3s^4+24nc^2s^2+46s^2c^2+2n^2s^3c+4ncs^3+8nc^4+n^2c^4-6cs^3+3n^2s^2c^2-6ns^4-56sc^3-2n^2s^4+16c^4-4n^2c^3s)b^2
    +(-2n^2c^4s+8s^5-24c^4s+2n^2s^3c^2+2s^5n+4s^3nc^2+16s^2nc^3-14c^4ns-6s^3c^2+32s^2c^3+2n^2s^2c^3-2n^2s^4c-8s^4nc-14cs^4)b
    -6s^3c^3+8s^5c+2s^5nc-2n^2s^4c^2-6s^4nc^2+n^2s^6-4s^6+6ns^2c^4+9s^2c^4+n^2s^2c^4-2s^3nc^3-3s^4c^2...................(3)

Let n=2, s=5, c=1, we obtain
v^2 = 361b^4-38b^3-11099b^2+16500b+22500................................................................................(4)

This quartic curve is birationally equivalent to an elliptic curve below.

Y^2 = X^3 -4636225X + 3673375625 with generator: p1=[950 ,-11725 ], p2=[1472 , 5501].

Hence we choose positive rational solution b of equation (4), and we need that a>0 and d>0.

From equation (2), a = 1/2(19b^2+300-151b+v)/(75-19b), d = 1/2(19b^2+300-151b-v)/(75-19b).

Thus a>0 and d >0 if 0 < b <1 or 3 < b <75/19.

Hence there are rational solutions a,d>0 with 0 < b <1 or 3 < b <75/19.

We show some solutions.

[a, b, c, d]
[207767, 24375, 68675, 42558]
[1239385342, 6969120375, 2302376995, 1001002263]
[1446725553, 25651538000, 8504601713, 6920143299]
[1152519233419530431168, 47210918250957408022625, 15695150257812104233483, 14417163546871478480139]
[924617488018326314112932040418355883, 97558196524088146810963579255531625, 1032497255384725888043730368567334953, 3107813336996489091251025854595452304]

Q.E.D.




3.Results

Search the rational solutions of quartic equation (3) for b with 3=< n <=30. 
s < 30000
c < s/4

Solutions are not always smallest.

[n][ a  b  c  d]

[3][59, 5, 200, 11]
[4][46280, 731, 4505, 7049]
[5][775, 26, 39, 96]
[6][1050125, 26553, 117130, 36057]
[7][744700, 57339, 21329, 29792]
[9][46816, 744, 1085, 3435]
[10][35707, 1612, 1512, 481]
[11][57455, 512, 1568, 3185]
[12][4351, 117, 131, 117]
[13][59415983, 2141759, 167453, 2287520]
[14][12705, 89, 534, 289]
[15][1301373, 268, 34540, 52327]
[16][13389, 124, 527, 189]
[17][66735985, 110791, 3044344, 783904]
[18][11128875, 238391, 117490, 264244]
[19][1425665, 2375, 15281, 57584]
[20][7601863, 81046, 82810, 217165]
[21][447069, 6125, 7435, 7776]
[22][5806561671, 70623490, 156594035, 37251084]
[23][54410387, 533922, 1034191, 801922]
[24][787989355, 5757289, 14338480, 12792856]
[25][716783760, 17571612, 2958334, 8186462]
[26][1238503, 1760, 2088, 43857]
[28][925216313, 33263, 19600694, 13451018]
[29][16987286, 25016, 248685, 312753]
[30][513520, 8034, 3605, 5497]



4.Reference

[1]: Andrew Bremner, Allan Macleod, An unusual cubic representation problem, Ann. Math. Inform. 43 (2014)