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

#NumberTheory #ConicSections #Competitions

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 $x_1,...x_n$ which satisfies $x_1>x_2>...>x_n$ , 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 years, 3 months ago

Are you really 14 years old?

Syed Hamza Khalid - 2 years, 8 months ago

I'm sure 10th graders are 14-15 years old, so yep, you're right.

Just a math lover :)

P/S: Not bragging, but these are still easy to what we have to do :) If you really want to screw your brain up, I have some for you :D

Steven Jim - 2 years, 8 months ago

@Steven Jim – I am in grade 10 and 99% of my classmates don't even know the quadratic equation or factorials. They only know BIDMAS and cubic graphs nothing else special.... hence I look like a genius at my class but thanks to Brilliant; I identified that it's the school's fault of having such low mathematic standards and that globally i am literally somewhat a zero hero :(

However, due to Brilliant; I am able to increase my standard and understanding of mathematics in a dramatic way.

Syed Hamza Khalid - 2 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$