Can you solve these Number Theory problems in 30 seconds & 1 minute?

If possible, list all the solutions in limited time too!

You have 30 seconds for solving these problems.

Def: Lattice point is the point \((x,y)\) such that both \(x\) and \(y\) are integers.

1.) Find the number of lattice points of a hyperbola given by an equation

\[4x^{2}-y^{2}-2y-2558 = 0\]

2.) Find the number of lattice points of an ellipse given by an equation

\[2x^{2}-xy+2y^{2}+3x-3y-3 = 0\]


You have 1 minute for solving these problems.

1.) Find the number of ordered pair \((x,y)\) in integers of the equation

\[x^{2} - y! = 2559\]

2.) Find the number of ordered triples \((x,y,z)\) in integers of the equation

\[x^{2}+y^{2}+z^{2} = 2558xyz\]


Sometimes the math competitions in my country are too goddamn crazy. I hate it, and I'll never have this competition again. XD

Note by Samuraiwarm Tsunayoshi
2 years, 10 months ago

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I just might say I can.

For the first question, we change it to \(4{ x }^{ 2 }-{ (y+1) }^{ 2 }=2557\), and as \(4x^2\) is divisible by 4, \(4{ x }^{ 2 }-{ (y+1) }^{ 2 }\) will have a remainder of 0 or -1 when divided by 4, which is not the case here.

Next question, multiply both sides by 2, then we will have \({ (x-y) }^{ 2 }+{ 3(x+1) }^{ 2 }+{ 3(y-1) }^{ 2 }=12\), and we can then claim that \({ (x-y) }^{ 2 }\) must be 0 or 9 (as 12 and \({ 3(x+1) }^{ 2 }+{ 3(y-1) }^{ 2 }\) are all divisible by 3). The last part is pretty lengthy, so I will not write it down here (besides, it's easy from here)

The third one, we have \({ x }^{ 2 }=2559+y!\), which means \(x^2\) will have the remainder of 3 when divided by 4 if \(y \ge 4\), which is impossible. Test for numbers from 1 to 3, tada, we have the answer.

The final one, if none of the 3 is divisible by 2 then \(x^2+y^2+z^2\) is not divisible by 2 while \(2558xyz\) is not. If one or two is then \(x^2+y^2+z^2\) is not divisible by 4 while \(2558xyz\) is, which means that all 3 must be divisible by 2. Let \({x}^{'}=\frac{x}{2},{y}^{'}=\frac{y}{2},{z}^{'}=\frac{z}{2}\) and prove exactly like shown above. Also note that if \(x,y,z\) are positive integers, then there does not exist a set of infinitely many numbers \(a_1,...,a_n\) which satisfies \(a_1>a_2>...>a_n\), which leaves \(x=y=z=0\) the only solution.

The first 2 problems I solved within 20 seconds each, while the third was solved in 10 seconds and the final one took me almost a minute. I think I should start celebrating now.

Steven Jim - 3 months, 2 weeks ago

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