Intereting Posts

Why are affine varieties except points not compact in the standard topology on $C^n$ ?
Problem 1.1.4 from “Geometry Revisited”
If 5 points are necessary to determine a conic, why are only 3 necessary to determine a parabola?
Largest Part of a Random Weak Composition
For what values of $n$ is $n^2+n+2$ a power of $2$?
How to compute the sum $ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$
Closed form for $\sum_{n=1}^\infty \left(e-\left(1+\frac{1}{n}\right)^n \right)^2$?
(Locally) sym., homogenous spaces and space forms
Intermediate value property problem and continuous function
Why is the graph of a continuous function to a Hausdorff space closed?
Size of a family of sets $F$ such that if $|A\cap X|=|B\cap X|$ for all $X\in F$, then $A=B$
Relation between total variation and absolute continuity
Intuitively, how should I think of Measurable Functions?
Do isomorphic quotient fields imply isomorphic rings?
Expected value of $x^t\Sigma x$ for multivariate normal distribution $N(0,\Sigma)$

Another Project Euler problem has me checking the internet again. Among other conditions, four of my variables satisfy:

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

According to Wikipedia, this is known as a Pythagorean Quadruple. It goes on to say all quadruples can be generated from an odd value of $a$ and an even value of $b$ as:

- Solving $x^p + y^p = p^z$ in positive integers $x,y,z$ and a prime $p$
- Number of Solutions to a Diophantine Equation
- On $p^2 + nq^2 = z^2,\;p^2 - nq^2 = t^2$ and the “congruent number problem”
- Numbers of the form $\frac{xyz}{x+y+z}$
- The Diophantine equation $x^2 + 2 = y^3$
- Total number of solutions of an equation

$$c=\frac{a^2+b^2-p^2}{2}, \quad d=\frac{a^2+b^2+p^2}{2} ,$$

where $p$ is any factor of $a^2+b^2$ that satisfies $p^2<a^2+b^2$.

However, I can’t see how or why this works. I also can’t seem to see how this works for $\lbrace 4,4,7,9 \rbrace$. Am I missing something here?

- Generalizing Ramanujan's 6-10-8 Identity
- Why does $x^2+47y^2 = z^5$ involve solvable quintics?
- Finding all the numbers that fit $x! + y! = z!$
- Finding primes so that $x^p+y^p=z^p$ is unsolvable in the p-adic units
- General formula to obtain triangular-square numbers
- Diophantine equation: $n^p+3^p=k^2$
- Quintic diophantine equation
- Link between the negative pell equation $x^2-dy^2=-1$ and a certain continued fraction
- The equation $x^n+y^n+z^n=u^n+v^n+w^n=p$, where $p$ is prime.
- How to solve $(2x^2-1)^2=2y^2 - 1$ in positive integers?

I think $c$ and $d$ should have been

$$

\begin{split}

c &= \frac{a^2+b^2-p^2}{2 p}\qquad\qquad

d &= \frac{a^2+b^2+p^2}{2 p}

\end{split}

$$

According to this document a set of Pythagorean Quadruples is given by:

$a=2mp$

$b=2np$

$c=p^2-(n^2+m^2)$

$d=p^2+(n^2+m^2)$

where $m,n,p$ are integers such that:

$m+n+p \equiv 1 (\mod 2) \land gcd(m,n,p)=1$

Sum3Squares

I thought of extremely simple derivation of the parametrization of three squares which sum to square.

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

then $a^2 + b^2 = d^2 – c^2$

As is well known, any number which is the sum of two squares is the product of only primes which are the sum of two squares.

We can thus easily derive from complex number arithmetic the parametrization of composite numbers which are the sum of two squares.

$$(a_1 + i b_1)(a_1-i b_1)(a_2 + i b_2)(a_2 – i b_2) = (a_1 + i b_1)(a_2 + i b_2) (a_1 – i b_1) (a_2 – i b_2)$$

$$(a_1^2 + b_1^2) (a_2^2 + b_2^2)

= ( (a_1 a_2 – b_1 b_2) + i (a_1 b_2 + b_1 a_2) )( (a_1 a_2 – b_1 b_2) – i (a_1 b_2 + b_1 a_2) )$$

$$(a_1^2 + b_1^2) (a_2^2 + b_2^2) = (a_1 a_2 – b_1 b_2)^2 + (a_1 b_2 + b_1 a_2)^2$$

Also, since the product of two numbers is the difference of the squares of half their sum and half their difference,

$$(a_1^2 + b_1^2)(a_2^2 + b_2^2)

= ((a_1^2 + b_1^2 + a_2^2 + b_2^2)/2)^2 – ((a_1^2 + b_1^2 – a_2^2 -b_2^2)/2)^2$$

Thus

$$(a_1 a_2 – b_1 b_2)^2 + (a_1 b_2 + b_1 a_2)^2

= ((a_1^2 + b_1^2 + a_2^2 + b_2^2)/2)^2 – ((a_1^2 + b_1^2 – a_2^2 -b_2^2)/2)^2$$

$$(a_1 a_2 – b_1 b_2)^2 + (a_1 b_2 + b_1 a_2)^2 + ((a_1^2 + b_1^2 – a_2^2 -b_2^2)/2)^2

= ((a_1^2 + b_1^2 + a_2^2 + b_2^2)/2)^2$$

$(a_1 a_2 – b_1 b_2)^2 + (a_1 b_2 + b_1 a_2)^2$ not have remainder 2 when divided by 4.

That is why it is required that not both $(a_1 a_2 – b_1 b_2)$ and $(a_1 b_2 + b_1 a_2)$

be odd.

Kermit

- To create a special matrix !!
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