1.Introduction


We show  X^4 + Y^4 + Z^4 + W^4 = T^2 has infinitely many parametric solutions.


2.Theorem
      
 
There are infinitely many parametric solutions of X^4+Y^4+Z^4+W^4 = T^2.
      
      2P: {X,Y,Z,W,T}={16*n^2*(n^2-3)*(n-1)*(n+1)*(n^8-4*n^6+46*n^4-52*n^2+25),
      
                       -2*n*(n^2-3)*(n^4-10*n^2+5)*(n^8-4*n^6+46*n^4-52*n^2+25),
                       
                       -(n^8-4*n^6+46*n^4-52*n^2+25)*(n-1)*(n+1)*(n^2+5)*(n^4-10*n^2+5),
                       
                       2*(n-1)*(n+1)*(n^4-10*n^2+5)*(n^6+27*n^4-13*n^2+25)*(n^2-3)*n,
                       
                       (n^16-2*n^15-8*n^14-26*n^13+108*n^12+686*n^11-472*n^10-2906*n^9+2582*n^8+4618*n^7-4984*n^6-4510*n^5+5004*n^4+2890*n^3-2600*n^2-750*n+625)
                       *(n^16+2*n^15-8*n^14+26*n^13+108*n^12-686*n^11-472*n^10+2906*n^9+2582*n^8-4618*n^7-4984*n^6+4510*n^5+5004*n^4-2890*n^3-2600*n^2+750*n+625)
                  }
      
      -2P: {X,Y,Z,W,T}={-16*n^2*(n^8+24*n^6+178*n^4-544*n^2+325)*(n-1)*(n+1)*(n^6+27*n^4-13*n^2+25)*(n^10+37*n^8+378*n^6-134*n^4-875*n^2+625)*(n^10+13*n^8+250*n^6-534*n^4+277*n^2+25),
      
                        -2*n*(n^8+24*n^6+178*n^4-544*n^2+325)*(n^10+33*n^8+410*n^6-686*n^4+85*n^2+125)*(n^10+37*n^8+378*n^6-134*n^4-875*n^2+625)*(n^10+13*n^8+250*n^6-534*n^4+277*n^2+25),
                        
                        -(n^10+13*n^8+250*n^6-534*n^4+277*n^2+25)*(n^10+37*n^8+378*n^6-134*n^4-875*n^2+625)*(n-1)*(n+1)*(n^2+5)*(n^6+11*n^4-93*n^2+25)*(n^10+33*n^8+410*n^6-686*n^4+85*n^2+125),
                        
                        2*(n-1)*(n+1)*(n^10+33*n^8+410*n^6-686*n^4+85*n^2+125)*(n^18+81*n^16+1636*n^14+6900*n^12-8514*n^10+253806*n^8-485612*n^6+255300*n^4-24375*n^2+15625)*(n^8+24*n^6+178*n^4-544*n^2+325)*n,
                        
                        (70567715000*n^9-606113565960*n^11+46635031250*n^7+4511718750*n^3-18236718750*n^5-1269531250*n+n^40+100*n^38+36702672*n^30+1429056144*n^26+319079976*n^28+4718*n^36+137892*n^34+2724285*n^32
                        -1139161006*n^24+2534296892*n^23-33787485832*n^22+16032070172*n^21-230146736380*n^19+1308574366376*n^13+823635892644*n^17-1416658739096*n^15+65585467476*n^20+86745936056*n^18+324693334824*n^12
                        +4767968750*n^4-359702739598*n^16+255789354768*n^14+4726562500*n^2+207467544*n^27+14990*n^35+437566*n^33+7256456*n^31-156671437500*n^6+2*n^39+274*n^37+63912648*n^29-344889128*n^25-708024116400*n^10
                        +514985518125*n^8+244140625)*(-70567715000*n^9+606113565960*n^11-46635031250*n^7-4511718750*n^3+18236718750*n^5+1269531250*n+n^40+100*n^38+36702672*n^30+1429056144*n^26+319079976*n^28+4718*n^36
                        +137892*n^34+2724285*n^32-1139161006*n^24-2534296892*n^23-33787485832*n^22-16032070172*n^21+230146736380*n^19-1308574366376*n^13-823635892644*n^17+1416658739096*n^15+65585467476*n^20+86745936056*n^18
                        +324693334824*n^12+4767968750*n^4-359702739598*n^16+255789354768*n^14+4726562500*n^2-207467544*n^27-14990*n^35-437566*n^33-7256456*n^31-156671437500*n^6-2*n^39-274*n^37-63912648*n^29+344889128*n^25
                        -708024116400*n^10+514985518125*n^8+244140625)
                     }
      n: arbitrary



     
Proof.

X^4+Y^4+Z^4+W^4 = T^2....................................................................................(1)

Set X=a, Y=b, Z=c, W=x, T=x-k............................................................................(2)

x^2 = 1/2(k^2-a^4-b^4-c^4)/k.

Set k=a^2+b^2+c^2, then 

x^2 = (a^2b^2+a^2c^2+b^2c^2)/(a^2+b^2+c^2)...............................................................(3)

Hence we have to find a rational solution of {u^2=a^2b^2+a^2c^2+b^2c^2, v^2=a^2+b^2+c^2}.

Fortunately, Euler had already found this solution below.(Piezas[1])

{a,b,c} = {p^2+q^2-r^2,  2pr,  2qr} where {p,q,r} = {-8n(n^2-1),  n^4-10n^2+5,  -n(n^2+1)(n^2-3)}.

Substitute Euler's solution to (3), and we obtain

x = 2(n-1)(n+1)(n^2+1)(n^4-10n^2+5)(n^6+27n^4-13n^2+25)(n^2-3)n/((n^8-4n^6+46n^4-52n^2+25)).

Substitute x to (2), and we obtain a solution of (1).(mP of m=2)


According to Euler[2], he obtained  the solution as follows.

Substitute a=2rp, b=2rq, and c=p^2+q^2-r^2 which is the solution of v^2=a^2+b^2+c^2 to u^2=a^2b^2+a^2c^2+b^2c^2,
then we obtain (4).

s^2=(q^2+p^2)r^4+(-2q^4-2p^4)r^2+3q^2p^4+3q^4p^2+q^6+p^6..................................................(4)
Substitute r=p-nq to (4), then we obtain 

t^2 = 4(1+n^2)p^4-4n(1+n^2)p^3q+(1+6n^2+n^4)p^2q^2+4n(1-n^2)pq^3+(1-n^2)^2q^4.............................(5)

Let transform equation (5) to elliptic curve (6).
X=p/q

Y^2-4nYX+(8n-8n^5)Y = X^3+(n^4+2n^2+1)X^2+(-16+16n^2+16n^4-16n^6)X-16n^10-16n^8+32n^6+32n^4-16n^2-16......(6)

Since P = (-n^4-2n^2-1, 4n^5-8n^3-12n) is a point on (6), we can obtain a point 2P and corresponding point 2Q as follows.

2Q=(-8n(n-1)(n+1)/(n^4-10n^2+5), (n-1)(n+1)(n^2+1)(n^6+27n^4-13n^2+25)/((n^4-10n^2+5)^2))

Since q/p=-8n(n-1)(n+1)/(n^4-10n^2+5), we obtain {p,q,r}= {-8n(n-1)(n+1), n^4-10n^2+5, -n(n^2-3)(1+n^2)}.

Euler gave the solution in this way.

Furthermore we can obtain the points mP on the elliptic curve (6), m=1,2,3,....,
then the equation (1) has infinitely many parametric solutions.

We show two parametric solutions of the case 2P and -2P. 


Q.E.D.
　



3.References
 
[1].Tito Piezas: {x^2+y^2+z^2, x^2y^2+x^2z^2+y^2z^2}
[2].L.Euler: On three square numbers, of which the sum and the sum of products two apiece will be a square