\hspace{-16}$(1): The no. of Integer ordered pairs $\bf{(x,y)}$ in $\bf{x^2+y^2 = 2013}$\\\\\\ $(2):$ The no. of Integer ordered pairs $\bf{(x,y,z)}$ in $\bf{\begin{Vmatrix} \bf{x=yz} \\ \bf{y=zx} \\ \bf{z=xy} \end{Vmatrix}}$\\\\\\ $(3):$ The no. of Integer ordered pairs $\bf{(x,y,z)}$ in $\bf{\begin{Vmatrix} \bf{x+yz = 1} \\\\ \bf{y+zx = 1} \\ \bf{z+xy = 1} \end{Vmatrix}}$\\\\
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2 Answers
2305(1)-Has been done before [See-"http://www.targetiit.com/discuss/topic/21861/no-of-solution/"]
Answer = 0
(2)- 4 [they are-(1,1,1),(-1,1,-1),(1,-1,-1),(-1,-1,1])
- Shaswata Roy ActuallY 5.I forgot the most trivial one-(0,0,0)Upvote·0· Reply ·2013-05-22 21:11:27
man111 singh Thanks Shaswata
1357(2) multiply all 3 to get xyz=1 and proceed
(3) subtract second from first to get x-y and z
- man111 singh Thanks manish Sir