x^2 + xy + y^2 = 3 and x + xy + y + 1 = 0?

Question

the graph of x^2 + xy + y^2 = 3 intersects the graph of x + xy + y + 1 = 0 in k distinct points. The value of k is .

Answer 1

I don't know, but here are a couple of ideas to start with:

1. At those points, what is the value of (x+y+1)^2?
2. Subtract the second equation from the first, and complete the squares in x and y.

Answer 2

Write x = A^mX, y = B^mY and z = C^mZ, the place all top powers p^e dividing X, Y or Z have exponent 0 < e < n. So x^(a million/n) + y (a million/n) = z^(a million/n) turns into AX^(a million/n) + by using^(a million/n) = CZ^(a million/n). additionally, with out loss of generality we are able to assume gcd(x,y,z) = a million. even however, i will assume that gcd(n,xyz) = a million and for each X and Y there exist an e top to n. Write ? = X^(a million/n), ? = Y^(a million/n), ? = Z^(a million/n), so as that ? ? ? are algebraic integers friendly the polynomials f(T) = T^n - X = 0, g(T) = T^n - Y = 0, and h(T) = T^n - Z = 0. each and each poly is irreducible over Q. enable e be the backside top exponent in X, then if f(T) = r(T)s(T), r(T)s(T) = f(T) = T^n mod p^e. So then r(T) = T^u and s(T) = T^v mod p^e, yet then the consistent term in the two r and s is divisible by using p^e, making the consistent term in f(T) divisible by using p^2e rather of in basic terms p^e. Adjoin the two root ? and ? to Q then F = Q(?, ?) and write R for the ring of algebraic integers in F. this is sparkling from the form of f(T) and g(T) that the only ramified primes are those dividing X and Y, and probably some dividing n (which we limit out latter). Taking nth powers of the two facets z = (C?)^n = (A? + B?)^n = x + (n,a million)(A?)^(n-a million)(B?) + ....+ (n,a million)(A?)(B?)^(n-a million) + y. Rewrite the equation as z-x-y = (A?)(B?)((n,a million)(A?)^(n-2) + ....+ (n,a million)(B?)^(n-2))(**). So z ? x + y mod A? and mod B?. all of us comprehend x is divisible by using A? and y is divisible by using B?, so if some top appropriate P in R divides the two A? and B? then it divides all 3 x, y and z, yet gcd(x,y,z) = a million, so gcd(?, ?) = a million. for the reason that gcd(?, ?) = a million and ? and ? define the ramifided primes then ok= Q(?)?Q(?) = Q for the reason that no top ramifies in ok (you'll be able to decide for some extra regulations on n for this, like if p top divides n then X and Y can not be p powers mod p). So the powers ?^i?^j type a foundation over Z for some subring of R. The left factor (**) is a rational integer and the amazing a in simple terms algebraic. the only way this is usually is that if the two facets = 0, bringing us to the trivial case. Sorry the only way i comprehend the thank you to try this is by using staring at how top ideals ingredient in the two extensions.