Question

A Pythagorean triple is a triple of natural numbers satisfying the equation a^2+b^2+c^2.

Answer 1

X, x+2 - two consecutive odd or even numbers
Add the reciprocals of these numbers.
$\frac{1}{x}+\frac{1}{x+2}=\frac{x+2}{x(x+2)}+\frac{x}{x(x+2)}=\frac{x+2+x}{x^2+2x}=\frac{2x+2}{x^2+2x}$

Now add the squares of the numerator and denominator, as in the example.
$(2x+2)^2+(x^2+2x)^2= \\ 4x^2+8x+4+x^4+4x^3+4x^2= \\ x^4+4x^3+8x^2+8x+4$

So this number has to be a perfect square.
$x^4+4x^3+8x^2+8x+4= \\ x^4+2x^3+2x^2+2x^3+4x^2+4x+2x^2+4x+4= \\ x^2(x^2+2x+2)+2x(x^2+2x+2)+2(x^2+2x+2)= \\ (x^2+2x+2)(x^2+2x+2)= \\ (x^2+2x+2)^2$
It is a perfect square, so this method always works.

The numbers $2x+2, \ x^2+2x, \ (x^2+2x+2)^2$ are a Pythagorean triple for any $x \in \mathbb{N^+}$.

Answer 2

Answer:

even tho this has nothing to do with the answer ;-;

Step-by-step explanation:First a definition: A Pythagorean Triple are three natural numbers 1 <= a <= b <= c, such that a2 + b2 = c2 holds. For example 3, 4, 5 is such a triple, since 32 + 42 = 9 + 16 = 25 = 52. While 2, 3, 4 is not such a triple, since 22 + 32 = 4 + 9 = 13 and 42 = 16. We note here that only natural numbers are considered, and thus 2, 3 can not be extended to Pythagorean triple (since 13 is not the square of some integer).

Now the question: Can we colour the natural numbers 1, 2, 3, ... with two colours, say blue and red, such that there is no monochromatic Pythagorean triple? In other words, is it possible to give every natural number one of the colours blue or red, such that for every Pythagorean triple a, b, c at least one of a, b, c is blue, and at least one of a, b, c is red ? We prove: The answer is No. That is easier to express positively: Whenever we colour the natural numbers blue or red, there must exist a monochromatic triple (one blue triple or one red triple).

More precisely we prove, using "bi-colouring" for colouring blue or red: 1) However we bi-colour the numbers 1, ..., 7825, there must exist a monochromatic Pythagorean triple. 2) While there exists a bi-colouring of 1, ..., 7824, such that no Pythagorean triple is monochromatic. Part 2) is relatively easy. Part 1) is the real hard thing -- every number from 1, ..., 7825 gets one of two possible colours, so altogether there are 27825 possible colourings, which all in a sense need to be considered, and need to be excluded. What is 27825? It is approximately 3.63 * 102355, that is, a number with 2356 decimal places. The number of particles in the universe is at most 10100, a tiny number with just 100 decimal places (in comparison).

Now let's perform real brute-force, running through all the possibilities, one after another: Even if we could place on every particle in the universe a super-computer, and they all would work perfectly together for the whole lifetime of the universe -- by far not enough. Even not if inside every particle we could place a whole universe. Even if each particle in the inner universe becomes again itself a universe, with every particle carrying a super-computer, still

by far not enough. Hope you get the idea -- the $100 we got wouldn't pay that energy bill.

Fortunately there comes SAT solving to the rescue, which actually is really good with such tasks -- it can solve some such task and even more monstrous tasks. Our ``brute-reasoning'' approach solved the problem and resulted into a 200 terabytes proof -- the largest math proof ever. Though we must emphasise that this is in no way guaranteed, and possibly it will take aeons! SAT solving uses propositional logic, in the special form of CNF (conjunctive normal form). Fortunately, in this case it is easy to represent our problem in this form.