Find all points in the x y -plane with integer coordinates located at a distance 5 from the orig

Find all points in the x y -plane with integer coordinates...

Key Concepts

Pythagorean Triples

Pythagorean triples are sets of three positive integers that satisfy the Pythagorean theorem. They are directly relevant in the context of circle problems where the radius is an integer, such as 5, because they provide a starting point for identifying potential integer solutions to the circle equation.

Lattice Points

Lattice points are points in the coordinate plane whose coordinates are both integers. They play a crucial role in number theory and geometry, especially when solving equations that require integer solutions, as seen in problems involving circles with integer radius and centers.

Euclidean Distance Formula

The Euclidean distance formula, derived from the Pythagorean theorem, allows us to calculate the distance between any two points in the plane. It is expressed as the square root of the sum of the squares of differences in each coordinate, and is fundamental when determining if a given point lies on a circle of a certain radius.

Circle Equation

The circle equation represents all points that are at a fixed distance, known as the radius, from a given point (the center). In problems requiring lattice points on a circle, the equation is set up by equating the sum of the squares of the differences in coordinates to the radius squared, leading to questions about finding integer solutions.

Diophantine Equations

Diophantine equations are equations that seek integer solutions. When the circle equation is set to equal the square of the radius, the resulting equation is often a Diophantine equation. Solving such equations involves finding all lattice points that satisfy the circle equation, highlighting the interplay between algebra and number theory.

Transcript

00:01 So for this question, we are trying to find which points on the graph with integer coordinates are five units away from the origin.

00:20 So, and then we have to find that for the point to three, which we'll do in a bit.

00:27 So for now, we want to see which points are five units away from.

00:35 From the origin.

00:37 So logically, the first step is to graph it.

00:42 And we know that these points already labeled here are five units from the origin.

00:48 So we have four points already.

00:49 We have, let's see, we have 0 5.

00:54 One clockwise, we have 5, then we have 0 negative 5, and negative 5.

01:05 0.

01:07 Okay, so those are four points first.

01:10 And then the next points that we want to find are, well, we want to find the points such that the squares of those integers, basically of those like whole number values, are equal to 5, or the squares are equal to 25 actually, 5 squared.

01:38 And it makes sense because of the pythagorean theorem, if you do a squared plus b squared equals c squared.

01:46 For these points, if there's always a zero, you're going to have five squared at least once plus zero, and that equals 25.

01:56 We want to make sure that we have the c squared equals a 25 for another combination of points within this circle.

02:06 So if you think about all the perfect squares up to five, we have c0 square, which would be 0.

02:15 You'd have 1, 4.

02:18 You'd have 9, 16, and those are all the perfect squares you can work with.

02:29 So the only pairs of numbers that end up adding to 25 are right here, 25 and 0, that's already up here.

02:40 And then 1 plus 16 that would be 17 that's not enough 4 plus 16 that would be 20 that's also not enough but 9 plus 16 is also 25 and these correspond to 3 and 4 so any combination of 3 and 4 will also have a distance of 5 so if we go 3 off here and 4 here there's 1 if you go 4 off here and three up people there another one same thing down here four minus three here and three minus four there minus three over there minus four three and likewise for that so all these points shown here create that circle with the radius of five which is essentially what we're trying to find um yeah so so in order to find it for the other point, two, three, two, three, we just have to kind of shift everything over to the right by two and up three...

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