$$ { ${a^2 + b^2 = c^2}$ } $$
$$ {} $$

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$$a^2 + b^2 = c^2$$

What type of triangle is this?

What did you expect a^2 + b^2 + c^2 (which is 144 + 169 + 12 ) to do? And how did it fail to do it? If c is the lon...
$$a^2 + b^2 = c^2$$

Vieta Jumping: Related to IMO problem 6, 1988: If $ab + 1$ divides $a^2 + b^2$ then $ab + 1$ cannot be a perfect square.

The famous IMO problem 6 states that if a,b are positive integers, such that ab + 1 divides a^2 + b^2 , then \frac{ a^2 + b^2}{ab + 1 } is a perfect square, namely, gcd(a,b)^2 . How about a modification of this problem: If a,b are (strictly) positive integers, such that ab + 1 divides a^2 + b^2 , then ab + 1 cannot be a perfect square. I am looking for a proof of the claim above, or a counter-example. One possible app...

What are the perfect squares for $a^2 + b^2$?

I have been writing a python program to generate integers that satisfy the perfect root of a^2 + b^2 over large ranges of values using brute force loops. Are there direct generating equations to solve this problem? Just to be clear, for example, a = 16 and b=30 form a perfect square, since \sqrt{16^2+30^2}=34 . So is a=1308 and b=3815 , and so on. In my searching, I read how Euclid solved the Diophantine equations for pythagor...
$$a^2 + b^2 = c^2$$

Why can't prime numbers satisfy the Pythagoras Theorem? That is, why can't a set of 3 prime numbers be a Pythagorean triplet?

Primes are all odd expect 2 , so if a, b, c don't contain 2 , a^2 + b^2 is even but c^2 is odd, then a^2 + b^2 = c^2 can't be true. Of course if c=2 , then a^2 + b^2 = c^2 can't be true. If a = 2 , then c>b then c^2 - b^2 \geq (b+2)^2 - b^2 = 4b + 4 > a^2 ...

Are there any positive integers $a, b, c, d$ such that both $(a, b, c)$ and $(b, c, d)$ are Pythagorean triples?

Pythagorean triple is a triple of integers (a, b, c) such that a^2+b^2=c^2 . Is there any Pythagorean triple such that, not only a^2+b^2 , but also b^2+c^2 is a square number? If not, how to prove it? I tried to prove non-existence the following way: If true, it would mean that there is a pair of integers such that both sum and difference of their squares is a square number. Let's call these integers a and b and a<b . Then...

Congruent number problem- one elementary statement

I'll prove that if three of the four integers \{\,a,b,a+b,a-b\,\} are squares, then the fourth is a congruent number. By the way, this result was first mentioned by Leonardo of Pisa, better known as Fibonacci. First, let's get a correct definition of congruent: a positive integer n is said to be congruent if there is a rational number x such that both of the numbers x^2\pm n are squares of rational numbers. Next, a couple of l...

Find a formula for all the points on the hyperbola $x^2 - y^2 = 1$? whose coordinates are rational numbers.

Let Y=a/b X=\sqrt { \frac {a^2 + b^2} {b^2}}=\frac 1 b \sqrt{a^2+b^2} Now, we just need a^2+b^2 to be a square, i.e a^2+b^2=c^2 be pythagorean triples. One nice parametrization is to let a=p^2-q^2, b=2pq , then a^2+b^2=(p^2+q^2)^2 , so X=\frac {p^2+q^2}{2pq} So (part of) our solutions are (X,Y)=\left(\frac {p^2+q^2}{2pq},\frac {p^2-q^2}{2pq}\right) . All the solutions can be found if you parametrize all the pythagorean ...

Pythagorean Theorem -- from Wolfram MathWorld

For a right triangle with legs a and b and hypotenuse c, a^2+b^2=c^2. (1) Many different proofs exist for this most fundamental of all geometric theorems. The theorem can also be generalized from a plane triangle to a trirectangular tetrahedron, in which case it is known as de Gua's theorem. The various proofs of the Pythagorean theorem all seem to require application of some version or consequence of the parallel postulate: proofs ...

Odd Pythagorian triplets

Suppose the numbers are a,b,c . Suppose a^2+b^2=c^2 . The order doesn't matter anyway. Since a and b are odd, a^2+b^2 will be even. But, c^2 is odd. Hence, no such triplet can exist....

Number theory

Number theory (or arithmetic or higher arithmetic in older usage) is a branch of pure mathematics devoted primarily to the study of the integers and integer-valued functions. German mathematician Carl Friedrich Gauss (1777–1855) said, "Mathematics is the queen of the sciences—and number theory is the queen of mathematics." Number theorists study prime numbers as well as the properties of objects made out of integers (for example, ration...

Proving that if $a,b,c\in \mathbb N$ and $a^2+b^2=c^2$ then $abc$ is even.

Let a,b,c\in \mathbb N and a^2+b^2=c^2 then abc is even. My attempt: If one or two numbers of a,b,c are even then we're done, so we'll have to show that at least one of them is even. Suppose a,b are odd, then a^2,b^2 are odd, so a^2+b^2 must be even. So c^2 is even then c is even. So abc is even. Is this enough to prove it?...

$x-\frac1x+y-\frac1y=4\,$ has no rational solutions?

Doesn't solve the problem, but I thought this was an interesting equivalence. Your question is equivalent to the the assertion: There are no integers a,b with b\neq 0 such that a^2+b^2 and a^2+2b^2 are both perfect squares. Part 1: A solution to your equation gives a solution to mine If x-\frac{1}{x}+y-\frac{1}{y}=4 for rational x,y , then there is a pair of nonzero integers a,b such that a^2+b^2 and a^2+2b^2 are perf...
$$a^2 + b^2 = c^2$$

There don't exist $a, b$ positive integers such that $a^2 + b^2$ and $a^2 - b^2$ are perfect squares

I need to prove that there don't exist a, b positive integers such that a^2 + b^2 and a^2 - b^2 are perfect squares. I suopose that a^2 + b^2 = c^2 and a^2 - b^2 = d^2 with c, d positive integers, so (a^2 + b^2)(a^2 - b^2) = c^2d^2 Therefore (a^2 + b^2)(a^2 - b^2) = (cd)^2 and a^4 - b^4 = (cd)^2 but this equation doesn't have solution in positive integers. Is that right?...

If $9\sin\theta+40\cos\theta=41$ then prove that $41\cos\theta=40$.

I noted that 9,40 and 41 are pythagorean triple, i.e 9^2+40^2=41^2 . This fact can be used to solve the question easily. I'll show the solution for the general case: a\cos\theta+b\sin\theta=c, given that a^2+b^2=c^2 a\cos\theta+b\sin\theta=c b\sin\theta=c-a\cos\theta Squaring both the sides: b^2\sin^2\theta=c^2+a^2\cos^2\theta-2ca\cos\theta Substituting c^2 with a^2+b^2 gives: b^2\sin^2\theta=a^2+b^2+a^2\cos^...

Pythagorean Theorem Proof Without Words (request for words)

Call "a" the short leg of any of the shaded triangles above, and "b" the long leg. Call "c" the hypotenuse. In the figure on the left, we see two small white squares within the larger square: one small square has sides equal to "a". The other has sides equal to "b". So the area in white on the left is a^2 + b^2 . In the figure on the right, we have rearranged the same 4 triangles to new positions with a larger square congruent to the f...
$$a^{2} + b^{2} = c^{2}$$

Proof: if $a^{2}+b^{2}=c^{2}$ and $(a,b,c)=1$, then $c$ is odd.

I have attempted using proof by contradiction, however, I keep getting lost in the details with my third case. I believe I need to use \bmod 4 but I'm not sure if it's okay. I'd appreciate pointers or how to improve the proof (thank you): Proof: Let a^{2} + b^{2} = c^{2} and (a, b, c) = 1 . Suppose that c is even and consider the following cases: Case 1: Let a and b be even. Then a^{2} + b^{2} being even \Rightarrow c^{2...
$$a^2 + b^2 = c^2$$

Let $a$, $b$, and $c$ be integers satisfying $a^2 + b^2 = c^2$. Prove: $abc$ must be even.

It can also be proven directly from a certain formula. The numbers a , b and c such that a^2 + b^2 = c^2 are called "Pythagorean triples." Given two integers m and n satisfying 0 < m < n , you can obtain a Pythagorean triple with this "recipe": a = n^2 - m^2 b = 2mn c = m^2 + n^2 So a and c could be odd, but b is even, and that's enough for abc to be even as well. For example: 5^2 + 12^2 = 13^2 and 5...
$$a^2+b^2 = c^2$$

Pythogorean Triples

Assume that a,b,c\geq 0 are integers, and that a^2+b^2 = c^2 . Note that if n is an even integer, then 4 divides n^2 . On the other hand, if n is odd, then n^2 leaves a remainder of 1 when divided by 4 . In total, if c is even, then 4 divides c^2 . If either a or b (or both) is odd, then 4 does not divide a^2+b^2 , as this sum will leave either a remainder of 1 or 2 when divided by 4 . Thus both a and ...
$$a^2 + b^2 = c^2$$

Why is $a^2 + b^2 = c^2$ (Linear algebra)

This definition of the norm holds for the dot product in Euclidean space. It does not feel like a axiom, because if we defined distance differently we clearly run into problems. It does not hold for curved space. The simple elementary explanation for this is the cutting of pieces of areas and rearrangement, but this might not work for calculating the distance in curved spaces. To give an example why this kind of argument which involves ...
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