A Nice Algebra Challenge - Olympiad Mathematics

A Nice Algebra Challenge - Olympiad Mathematics

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[masaaki]

もっと簡単にできます😮
2式よりxy=37-x^2-y^2
これを1式に代入してxyを消去
6x^2+7y^2=159 (3式とする)
合同式mod3において
6x^2≡0

159≡0

2x^2+21z^2=53 (4式とおく)

ただし、z=0即ちy=0の場合、1式においてx^2=-26 2式においてx^2=37 という矛盾を生じるのでz≠0即ちy≠0
したがってz=±1即ちy=±3
これを、1式と2式に代入してそれぞれ
(x, y)=(±4, ±3)(±11, ±3) 1式
(x, y)=(±4, ±3)(-+7, +-3) 2式
となるが共通の解を本問題の解とし、
(x, y)=(±4, ±3)

[wesleysuen]

The first two minutes can be simply regarded as dividing the given equations (first by second, on each side)!!! We just need to make sure that there’s no division by zero which is the case here.

[prabhudasmandal]

Good problem, better solution.I subtracted the 2nd from the 1st equn and easily got a quadratic equn y^2--6xy+63=0 where a=1, b=°°--6x, and c=63 and found x=4, y=3 and x=--4, y=--3(leaving +-21)

[kata]

هل السؤال حل المعادلتين أم حل الجملة...وبالتالي ذهبت الحلاوة

[rudrapatel]

Again a tricky solution
Very nice solution of problem and also an observation.☺️