A Nice System of Diophantine Equations | Math Olympiad

A Nice System of Diophantine Equations | Math Olympiad
Welcome to infyGyan ! In this video, we dive into a captivating system of Diophantine equations that's perfect for sharpening your problem-solving skills. Join us as we break down each step of the solution, providing clear explanations and helpful tips along the way. Whether you're gearing up for a Math Olympiad or just enjoy tackling challenging algebra problems, this video is for you. Can you solve it? Give it a try and let us know in the comments!
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System of equations
Diophantine Equations
Algebra
Math Olympiad
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Algebraic identities
Algebraic manipulations
Solving systems of equations
Solving cubic equation
Substitution
Factorization
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[@woobjun2582]

With the given x³ +y³ = 9 x² +y² = 5 Utilizing properties as (x +y)³ = 9 +3xy(x+y) (x +y)² = 5 +2xy Letting x +y =m and xy =n, m³ = 9 +3mn ...(eq1) m² = 5 +2n ...(eq2) Substituting eq2 into eq1, m³ -15m +18 =0; (m -3)(m² +3m -6) =0, that is, m -3 = 0; m = 3 where m² +3m -6 =0 produces non integers. By m =3 into eq.2 yields n =2, which is x +y =3 and xy =2. Then by the given, x² +y² = 5; (x² +y²)² = 5²; x⁴ +y⁴ +2x²y² = 25; x⁴ +y⁴ = 25 -2x²y² Now with x³ +y³ = 9 (x³ +y³)(x⁴ +y⁴) = 9(25 -2x²y²); x⁷ +y⁷ +x³y⁴ +x⁴y³ = 225 -18x²y²; x⁷ +y⁷ +x³y³(x+y) = 225 -18x²y²; x⁷ +y⁷ = 225 -18x²y² - x³y³(x+y) Thus, x⁷ +y⁷ = 225 -18•2² -2³•3 = 225 -18•4 -8•3 = 225 -72 -24; x⁷ +y⁷ = 129

[@johnstanley5692]

g1=x^2+y^2-5, g2=x^3+y^3-9, g3=x^7+y^7; g2/g1= (5-y^2)*x+y^3-9=> xo= (y^3 - 9)/(y^2 - 5) substitute xo into g1 (=0) => 2*y^6 - 15*y^4 - 18*y^3 + 75*y^2 - 44 = 0 -> -(y - 1)*(y - 2)*(- 2*y^4 - 6*y^3 + y^2 + 33*y + 22) Numerically, 3rd term factors with two real values :2*(y - 2.11064)*(y + 0.7383)*(y^2 + 4.3723*y + 7.0584) Real values: y={1, 2, -0.7384, 2.1106}=> x={2, 1, 2.1106, -0.7384}=>x^7+y^7={129.0000 129.0000 186.4777 186.4777}

[@ald6980]

Integer restriction to variables makes the problem trivial. 5=2^2+1^2 - the only decomposition of 5 as a summ of two squares. 2^3+1^3=9 that is (2,1) =(x,y) and there is no more solutions [if we change 2 and/or 1 to the opposite negative value we will decrease summ below than 9].

[@paulortega5317]

Same idea. Just no need to define a variable for xy. Let x + y = c 5 = x^2 + y^2 = (x + y)^2 - 2xy = c^2 - 2xy, or xy = (c^2 - 5)/2 9 = x^3 + y^3 = (x + y)^3 - 3xy(x + y) = c^3 - 3c(c^2-5)/2, or c^3 - 15c + 18 = 0 = (c -3)(c^2 + 3c - 6) c = 3 x + y = 3 (x,y) = (1,2) or (2,1) x^7 + y^7 = 129

[@kassuskassus6263]

x^7+y^7=129

[@TheAZZA0990]

20 secs in my head ..... 129!!! :)

[@RealQinnMalloryu4]

(3)+(6)= 9 3^2:(y ➖ 3x+3) .(2)+(3) (y ➖ 3x+2). {x^7+x^7 ➖}=x^14 {y^7+^y^7 ➖ {x^14+y^14}=xy^28 xy^2^14 xy^2^2^7 xy^1^2^1 xy^2^1 (y ➖ 2x+1).