مسلمانوں کا نیوٹن ۔ ڈاکٹر عبدالسلام

مسلمانوں کا نیوٹن ۔ ڈاکٹر عبدالسلام — Page 372

A Problem of Ramanujar۔By Abdus Salam Reprinted from the Maths, Student—Vol۔XI۔Nos۔1-2, Mar۔-June 1943 A Problem of Ramanujam Solve 2=a+y y=a+z z-a+u 1³-a+I (i) (ii) · (iii) (iv) 1۔Supposer, y, z, u are the roots of a biquadratic (* + p₂ t³ + p₂ l² + Pq ! + P₁ = 0۔We denote Σr" by S۔۔Now S₁ Pr S₁ = 4a + S₁ = 4α-p from the given equations Alan S۔+ P₁ S₁ + 2p = 0 Substituting for S₁, S, we have p۔- (a) B+P-40 2 (b) 2۔Subtract (iii) from (i) and (iv) from (ii) Fª³ - ¿' = y - u; y² - u² = z − x " (2ª³ − zª³){µ³ – u³) = (y − u}( z − x) or ±Y+2y+ux+ 112 − − 1 But Σxy = pa So ±z+uy-P₂+1 3۔arzy y-ay+yz Adding S₁-αS₁+xy+z! + ux:+12 ---1 Also ¹³-a²+au+ay + uy y³u² = a³ + az+01+2x Adding, ±²²+³u² = &a³ + α8, + n}} 4° zx (T2+y)=2a + aS₁ + p₂+1+2p۔(p₂+1)-2a²- np₁ + pa +1 +2p۔1۔2p,= p²² + p₂ − 2a² + api۔(c) (d) [zyzu-p₁] (e) 4۔Evidently or - (z³ − y³)(y³ − z³)(zª − u²){u² — z³) — (a: − y) − 2)(z - µ)(µ − x') (x+y)(y + z)(z+ «){µ+ x} = 1 or! (22}+u*y*+2ryzu) + Σ x²yz - 1 ( P₂ + 1)² + P₁ Ps – 4p۔— 1 Substituting for p, from (e) P₁ Pa − P₂² + 4a² – 2ap, – 0 {Σx³yz = p Ps - 4p,) (f) + Vide Collected Papers of Srinivasa Ramanujan, 1927, p۔332 Q۔722 Also J۔I۔M۔S۔Series I Vol VII p۔240۔362