Olympiad Level Algebra Challenge | Russian Math Olympiad Problem | Algebra Math | Advanced Math

Olympiad Level Algebra Challenge | Russian Math Olympiad Problem | Algebra Math | Advanced Math

algebra math, advanced math, olympiad exam, hard math questions, international mathematics olympiad, imo exam, international maths olympiad

[kienthanhle]

Kinda easy, this is my solution:
Notice how we can swap a and b and the equation stays the same, therefore, we set S = a+b, P = ab
We have
a^3 + b^3 = (a+b)^3 - 3ab(a+b) = S^3 - 3SP
a^2 + b^2 + ab + a + b = (a+b)^2 - ab + a + b = S^2 - P + S
We have a new system:
S^3 - 3SP = 7 <=> S^3 - 3SP - 7 = 0
S^2 - P + S = 4 <=> S^2 + S - 4 = P

Substitute P = S^2 + S - 4 into the first equation, we have:
S^3 - 3S(S^2 + S - 4) - 7 = S^3 - 3S^3 - 3S^2 + 12S - 7 = -2S^3 - 3S^2 + 12S - 7 = 0

<=> 2S^3 + 3S^2 - 12S + 7 = 0
We can easily guess S = 1 is a solution, therefore we factor S-1 out
(S-1)(S^2+5S-7) = 0
Therefore S = 1 or S^2+5S -7 = 0
Then we can easily figure out P, then we can easily figure out the solution

[oahuhawaii]

a^3 + b^3 = 7
a^2 + b^2 + a + b + a*b = 4
Symbols a and b are interchangeable.

The following equation has a and b as its roots:
(x - a)*(x - b) = 0
x^2 - (a + b)*x + a*b = 0
x^2 - Σ*x + Π = 0, where Σ = a + b, and Π = a*b
With the quadratic formula, we have:
a, b = (Σ ± sqrt(Σ^2 - 4*Π))/2

Σ^2 = (a + b)^2
= a^2 + b^2 + 2*a*b
= a^2 + b^2 + 2*Π
Σ^2 - 2*Π = a^2 + b^2

Equation 2:
a^2 + b^2 + a + b + a*b = 4
Σ^2 - 2*Π + Σ + Π = 4
Σ^2 - Π + Σ = 4
Σ^2 + Σ - 4 = Π

Equation 1:
a^3 + b^3 = 7
(a + b)*(a^2 - a*b + b^2) = 7
Σ*(Σ^2 - 2*Π - Π) = 7
Σ*(Σ^2 - 3*Π) = 7
Σ*(Σ^2 - 3*(Σ^2 + Σ - 4)) = 7
Σ*(-2*Σ^2 - 3*Σ + 12) = 7
0 = 2*Σ^3 + 3*Σ^2 - 12*Σ + 7
For Σ = 1, 0 = 2 + 3 - 12 + 7
(Σ - 1)*(2*Σ^2 + 5*Σ - 7) = 0
(Σ - 1)*(Σ - 1)*(2*Σ + 7) = 0
(Σ - 1)^2*(2*Σ + 7) = 0
Σ = 1, -7/2

Equation 2:
Π = Σ^2 + Σ - 4
= 1^2 + 1 - 4, (-7/2)^2 + (-7/2) - 4
= 1 + 1 - 4, 49/4 - 14/4 - 16/4
= -2, 19/4

For Σ = 1, Π = -2:
a, b = (1 ± sqrt(1^2 - 4*(-2)))/2
= (1 ± sqrt(1 + 8))/2
= (1 ± 3)/2
= 2, -1

For Σ = -7/2, Π = 19/4:
a, b = (-7/2 ± sqrt((-7/2)^2 - 4*19/4))/2
= (-7 ± sqrt((-7)^2 - 4*19))/4
= (-7 ± sqrt(49 - 57))/4
= (-7 ± i*2sqrt(2))/4 { discard complex solutions }

Thus, a = 2 and b = -1, or a = -1 and b = 2 .

[johnstanley]

Can solve this one by inspection - add 1 to both sides of 2nd equation to obtain:
a^2+b^2+(a+1)(b+1)=5 now choose b=-1 (or 'a') and equations become
a^3=8 & a^2=4 hence a=2, b=-1, or a=-1, b=2

[fouadhammout]

Saraha ma3andich zhar la f tssahib la f zwaj walit tangoul tawahad maynawad tawahda

[prabhudasmandal]

.... X^2+x_y-4=0 can be treated as quadratic equation where a=1, b=1 and c=_(y+4) giving x=2, and y=2+ves.thereafter no clue to find the result. Is it ok?

[MyOneFiftiethOfADollar]

Thanks for problem. Do you know what year for this particular Russian math Olympiad?

Also do not blame you for asking only for real solutions. Complex solutions were messy. The "guess and check" types would not have had much luck with them :)

[MacInTheNet]

I just guessed, then I skipped to the end and checked lol
Now I'm gonna watch it and if there are 10 minutes out there... it should mean it's quite difficult usually XD

[mathfreak]

It was kinda easy but I made it look complicated.

[opgamerz]

I used the logic that as 7 is its factors must be 1and 7 and solved i got the coreect ans