N+n²-n⁴+n²-y⁴
I need the value of x in factorise form
someone please help

Recall the equation x²+y²=z² from your school days. We associate this with Pythagoras’ theorem for calculating the sides of a right-angled triangle.

Mathematicians are also interested in natural number (positive integer) solutions to this equation. Such solutions are called Pythagorean Triples.

An example of a Pythagorean Triple is x=3, y=4, z=5 since 3²+4²=5². We can write it more conveniently as (3, 4, 5).

From any Pythagorean Triple, we can construct infinitely many Pythagorean Triples by multiplying through by a number e.g. multiplying (3, 4, 5) by 2 we get (6, 8, 10) which is also a Pythagorean Triple. An exercise for the reader is to show why this always works.

A primitive Pythagorean Triple (x, y, z) is one where x, y, and z are coprime, that is, they have no common factor greater than 1. (3, 4, 5) is primitive.

One challenge is to create a formula for generating Pythagorean Triples.

Here is Euclid’s Formula, where m and n can be any integers of our choice (providing m>n>0) and x, y and z will yield a Pythagorean Triple.

x=2mn

y=m²-n²

z=m²+n²

Example: If m=5 and n=1 then we get the Pythagorean Triple (10, 24, 26). Observe 10²+24²=100+576=676=26².

The following quick check shows that Euclid’s Formula will always work:

x²+y²=4m²n²+(m²-n²)²=4m²n²+m⁴+n⁴-2m²n²=m⁴+n⁴+2m²n²=(m²+n²)²=z²

However, it’s unsatisfying to be given this without understanding where it comes from. Below we will go through a method for deriving Euclid’s Formula from scratch:

Let’s rewrite x²+y²=z² as x²=z²-y²=(z-y)(z+y). This implies x/(z-y)=(z+y)/x.

Note: Division by zero is not an issue because if z=y then this would imply x=0 which is not allowed for a Pythagorean Triple.

However, given x, y and z are positive integers, x/(z-y) is some rational number. Recall the definition of a rational number is any number that can be expressed as m/n where m and n≠0 are integers e.g. 0.5 is a rational number since it’s equal to 1/2.

We can then set x/(z-y)=(z+y)/x=m/n for some integers m and n≠0. We then have the following two equations:

(z-y)/x=n/m — here we took the reciprocal

(z+y)/x=m/n

If we add equations 1 and 2 we get 2z/x=n/m+m/n=(n²+m²)/mn which means we can set x=2mn and z=n²+m².

Similarly, if we subtract equation 2 from 1 then we get 2y/x=m/n-n/m=(m²-n²)/mn which means we can set y=m²-n².

We have then arrived at Euclid’s formula for generating Pythagorean Triples (where m>n>0):

x=2mn

y=m²-n²

z=m²+n²

formula for generating Pythagorean Triples (where m>n>0):

x=2mn

y=m²-n²

z=m²+n²

i can't see the picture : /

2 & 7

step-by-step explanation:

he climbs 300 ft/hr at a constant rate. the peak is at 1,500 ft.

when going up, he reaches 600 ft in 2 hours. remember, that he is going in a constant rate. divide 600 with 300:

600/300 = 2

assuming that he takes the same route back (meaning that it's a 1500 ft route back, you can solve.

note also that you are solving when he gets back to 600 ft after he reaches the peak.

first, find the amount of hours it takes to get up the peak. divide the total amount (1500), with the ft/her (300) to get the amount of hours (x):

1500/300 = x

x = 5

it takes jamie 5 hours to get up. now, he is returning down, and reaches 600 (his current destination). remember that he climbs at a constant 300 ft/hr. divide 600 with 300:

600/300 = x

x = 2

now add the two numbers together.

x = 5 + 2 = 7

jamie takes 7 hours to go up the mountain and reach his current position when descending.

(total amount/constant rate) = (time).

remember that it is a constant rate, so in total, when he gets back down the mountain, it will take the same amount of time as when he goes up. the total amount for the round trip is: 5 + 5 = 10.

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