Difficult systems

Algebra Level 4

{ x 2 + y 2 = 61 x 3 − y 3 = 91 \begin{cases} { x }^{ 2 }+{ y }^{ 2 }=61 \\ { x }^{ 3 }-{ y }^{ 3 }=91 \end{cases}

Find sum of all solutions of the given equations above.

0 47 -i(47^(1/2)+74^(1/2)),or-i(47^(1/2)+74^(1/2)) 74 1 i^2 121

Shubham Haldēr by Shubham Haldēr , 22, India

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2 solutions

Ossama Ismail
Jan 12, 2021

The points of intersections are (5,6) and (-6,-5)

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x^2+y^2=k^2 has a solution only if k is a prime and of form 4k+1.. so as 61 is a prime and of 4k+1 its has solution x^2+y^2=61 .. it has 2 positive solutions and 8 integral solution .. x,y=5,6 ; 6,5 -5,6; 6,-5 5,-6; -6,5 -5,-6 ; -6 ,-5 x^3-y^3=91 every cube is of form 8k ,8k+/-1 ,8k+/-3 so 91=81k+3 form so 8k-(8k-3) or 8k+8k+3 are the only possibilities so x,y= 6,5 ; -6,-5 = 4,-3 ; 3,-4 are only solutions .. now intersection of both conditions give has (x,y)=(6,5) n (-6,-5) as only solutions .. so sum of all solutions possible = 6+5-6-5=0

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