# 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.

- 3 months, 2 weeks ago