Pythagoras' theorem is a special case of the Cosine theorem for a angle of $90°$. But also for an angle of 60° and 120°, "aesthetical" special cases derive:
$c^2=a^2+b^2\pm ab$
First question:
Are there further angle $x°$ with a rational number $x$, so that $\cos x$ is rational as well, thus creating "aesthetical" special cases?
Second question:
Does anybody know some internet sources to the equivalent of pythagorean triplets (Integers $a,b,c$, so that $c^2=a^2+b^2\pm ab$)?
If $\cos\frac {2m\pi}n$ is rational with $\gcd(m,n)=1$, then the primitive $n$th root of unity $\zeta=\cos \frac{2m\pi}{n}+i\sin\frac{2m\pi}n$ is a root of the rational polynomial $X^2-2\cos \frac{2m\pi}{n}X+1$. On the other hand, we know that $\zeta$ is a root of $X^n-1$ and any factorization of this over the rationals can be resaled to a factorization over the integers. Since $X^2-2\cos \frac{2m\pi}{n}X+1$ is monic this implies that $2\cos\frac{2m\pi }{n}$ is already an integer of absolute value $\le 2$. This leads to the cases
and that's all with $\gamma \in[0,2\pi)\cap \pi\mathbb Q$.
Regarding your second question: Pythagorean triples are best viewed as numbers $z=a+bi\in\mathbb Z[i]$ with norm $z\bar z$ a perfect square, which leads to a partitioning of the set of primes into those with $p\equiv -1\pmod 4$ (which can only occur as factor of $c$ if they are also factors of $a$ and $b$) and those with $p\equiv 1\pmod 4$ (which can be written as sums of squares and lead to primitive pythagorean triangles; for example $5=2^2+1^2=(2+i)(2-i)$ give us $\color{red}5^2=(2+i)^2(2-i)^2=(\color{red}3+\color{red}4i)(3-4i)$ and hence the most famous Pythagorean triangle) and the special prime $2$. Similarly, the numbers aou are after should be viewed as elements of $\mathbb Z[\omega]$ where $\omega=-\frac12+\frac i2\sqrt 3$. It turns out that one gets a similar partitioning of the primes, this time based on the remainders modulo $3$. Those are very interesting number theoretic questions indeed.
Regarding your first question:
The law of cosines states that
$$c^2=a^2+b^2-2ab\cos(\gamma)$$
If $r$ is any rational between $-1$ and $1$ you can set $\gamma=\arccos(r)$, therefore giving you
$$c^2=a^2+b^2-2abr$$
However the angle $\gamma$ is not a rational multiple of $\pi$ unless $r=0,\pm 1/2,\pm 1$. The latter is known as Niven's theorem.
As shown in this answer, the only rational values of $\cos\left(\frac pq\pi\right)$ for integer $p$ and $q$, are $\{-1,-\frac12,0,\frac12,1\}$. Thus, to get a rational cosine for rational degree angles we need an angle of $0^\circ$, $60^\circ$, $90^\circ$, or $120^\circ$ mod $180^\circ$
I have translated an article I posted on sci.math on what I called the skew Pythagorean triples.
Skew $\mathbf{-\frac12}$ Pythagorean Triples
Suppose that $a$, $b$, and $c$ are coprime positive integers so that $$ a^2+ab+b^2=c^2\tag{1} $$ Note that a triangle with these sides is obtuse and the cosine of the angle opposite $c$ is $-\frac12$. Therefore, let us call $(a,b,c)$ a skew $-\frac12$ Pythagorean triple.
Since $a$, $b$, and $c$ are coprime, at least one of $a$ or $b$ must be odd. Let us assume that $a$ is odd. If $b$ is odd, then $c^2$, being the sum of three odd numbers in $(1)$, is odd. If $b$ is even, then $c^2$, being the sum of one odd and two even numbers in $(1)$, is odd. Thus, $c$ is odd.
Since $a^2\equiv c^2\equiv1\pmod{8}$, $(a+b)b=ab + b^2\equiv0\pmod{8}$ by $(1)$. If $b$ is odd, then $a+b\equiv0\pmod{8}$. If $b$ is even, then $a+b$ is odd and $b\equiv0\pmod{8}$.
Therefore, we have two cases: $a$ and $b$ are both odd and $8$ divides $a+b$, or one is odd and $8$ divides the other.
Necessary Condition
Equation $(1)$ is equivalent to $$ 3a^2=(2c+2b+a)(2c-2b-a)\tag{2} $$ Let $d=\gcd(2c+2b+a,2c-2b-a)$. Then by $(2)$, we have $d^2\mid3a^2$; therefore, we also have $$ \begin{array}{lr} d\mid a&\qquad&\because d^2\mid3a^2\\ d\mid 4b&\qquad&\because4b=(2c+2b+a)-(2c-2b-a)-2a\\ d\mid 4c&\qquad&\because4c=(2c+2b+a)+(2c-2b-a) \end{array} $$ Thus, $d\mid\gcd(a,4b,4c)$. If $a$ is odd, $\gcd(a,4b,4c)=1$; therefore, $d=1$. If $a\equiv0\pmod{8}$, then $\gcd(a,4b,4c)=4$. Furthermore, both $b$ and $c$ are odd, so $4\mid2c+2b+a$ and $4\mid2c-2b-a$. Therefore, $d=4$. Notice that this means that $3$ divides only one of the factors in $(2)$, and it divides that factor an odd number of times.
For any coprime triple satisfying $(1)$, $(2)$ and the reasoning above says that $3a^2=3m^2n^2$ where $3$ does not divide $n$ and either $mn$ is odd and $gcd(m,n)=1$ ($a$ is odd), or $8$ divides $mn$ and $\gcd(m,n) = 2$ ($8$ divides $a$), and furthermore, $3m^2+n^2=4c$ and $|3m^2-n^2|=4b+2a$.
Therefore, for any coprime triple $(a,b,c)$ satisfying $(1)$, there exist $m$ and $n$ so that $3$ does not divide $n$ and either $mn$ is odd and $\gcd(m,n)=1$, or $8$ divides $mn$ and $\gcd(m,n) = 2$, so that $$ \begin{align} a&=mn\tag{3a}\\ b&=\frac{|3m^2-n^2|-2mn}{4}\tag{3b}\\ c&=\frac{3m^2+n^2}{4}\tag{3c} \end{align} $$ Conjugate Pairs
Since $3m^2-n^2-2mn=(3m+n)(m-n)$ and $n^2-3m^2-2mn=(n-3m)(n+m)$, to make $b > 0$ in $(3b)$, we need either $m\gt n$ or $n\gt 3m$. It turns out that for a pair $(m,n)$ where $m\gt n$, there is a pair $(m',n')$ where $n'\gt3m'$ for which $(a',b',c')=(b,a,c)$. In particular, $$ (m',n')=(m,n)\frac12\left[\begin{array}{r}1&3\\-1&1\end{array}\right]\tag{4a} $$ and $$ (m,n)=(m',n')\frac12\left[\begin{array}{r}1&-3\\1&1\end{array}\right]\tag{4b} $$ Thus, we only need consider one case or the other, for example, $m\gt n$, and then we can remove the absolute value from $(3b)$.
Thus, we have shown
Theorem 1:
All coprime triples $(a,b,c)$ so that $a^2+ab+b^2=c^2$ can be enumerated, without duplication, by taking two positive integers $m\gt n$, where $3$ does not divide $n$, and either $mn$ is odd and $\gcd(m,n)=1$, or $8$ divides $mn$ and $\gcd(m,n)=2$, and by setting $$ \begin{align} a&=mn\tag{5a}\\ b&=\frac{(3m+n)(m-n)}{4}\tag{5b}\\ c&=\frac{3m^2+n^2}{4}\tag{5c}\\ \end{align} $$
Note that $(4)$ and $(5)$ justify the claim above that for $n'\gt 3m'$, $$ \begin{align} a'&=b=n'm'\tag{6a}\\ b'&=a=\frac{(n'+m')(n'-3m')}{4}\tag{6b}\\ c'&=c=\frac{n'^2+3m'^2}{4}\tag{6c} \end{align} $$ Skew $\mathbf{\frac12}$ Pythagorean Triples
By the same reasoning as above, a triple $(a,b,c)$ which satisfies $$ a^2-ab+b^2=c^2\tag{7} $$ is a skew $\frac12$ Pythagorean triple. A triangle whose sides satisfy $(7)$ is not necessarily obtuse or acute; such triangles come in conjugate pairs, one acute and one obtuse. Let $(a,b,c)$ satisfy $(7)$ with $a\gt b$, then the triple $(a,a-b,c)$ also satisfies $(7)$. Furthermore, if $2b\gt a\gt b$, then $(a,b,c)$ is acute and $(a,a-b,c)$ is obtuse. If $a\gt2b$, then $(a,b,c)$ is obtuse and $(a,a-b,c)$ is acute.
A Triplet of Triples
If $(a,b,c)$ satisfies $(1)$ with $a\gt b$, then both $(a+b,b,c)$ and $(a,a+b,c)$ satisfy $(7)$; $(a+b,b,c)$ is obtuse and $(a,a+b,c)$ is acute. This is the conjugate pair of triples mentioned above.
Furthermore, if $(a,b,c)$ satisfies $(7)$ with $a\gt b$, then not only does $(a,a-b,c)$ satisfy $(7)$ as mentioned above, but $(a-b,b,c)$ satisfies $(1)$.
Therefore, each skew $-\frac12$ triangle is associated with two conjugate skew $\frac12$ triangles, one obtuse and one acute.
Thus, we have
Theorem 2:
All coprime skew $\frac12$ Pythagorean triples can be enumerated without duplication by $$ (a+b,b,c)\tag{8a} $$ which is acute if $a\lt b$ and obtuse if $a\gt b$, and $$ (a,a+b,c)\tag{8b} $$ which is obtuse if $a\lt b$ and acute if $a\gt b$, where $(a,b,c)$ is a skew $-\frac12$ Pythagorean triple as enumerated in Theorem 1.
To find all integer triples $a,b,c$ for which $c^2=a^2+b^2\pm ab$, it's enough to consider only $c^2=a^2+b^2+ab$ because changing the sign of $a$ or $b$ gives a solution to the other equation and vice versa.
We'll only look for primitive solutions, i.e, satisfying $\gcd(a,b,c)=1$. Then at least one of $a$ or $b$ is odd, say it's $b$.
Note that the equation can be rewritten $$4c^2=(2a+b)^2+3b^2.$$ Set $d=2a+b$, so $d$ is odd. We have $$3b^2=(2c-d)(2c+d)$$ where $\gcd(2c-d,2c+d)=1$ because of the primitivity. There are two cases: $2c-d=3u^2$ and $2c+d=v^2$, or $2c-d=v^2$ and $2c+d=3u^2$. In any case we have $\gcd(u,v)=1$, $3\nmid v$ and $b=uv$, with $u$ and $v$ odd.
The first case gives $c=\frac{v^2+3u^2}4$ and $d=\frac{v^2-3u^2}2$ from which $a=\frac{v^2-2uv-3u^2}4$.
The second case yields $c=\frac{v^2+3u^2}4$ and $d=\frac{3u^2-v^2}2$ from which $a=\frac{3u^2-2uv-v^2}4$.
In both cases $b=uv$.
As said before, we chose $b$ to be odd. This means we can change the values found for $a$ and $b$. (We could be introducing doubles here in the case where $a$ and $b$ are both odd. In fact, I'm sure we do.)
To obtain all values (including the non-primitive) we simply have to multiply by an integer constant $k=\gcd(a,b,c)$.
Example
Take coprime odd integers $u$ and $v$ with $3\nmid v$. Say $u=3$ and $v=5$. Then with the first case we obtain $a=-8$, $b=15$ and $c=13$. Indeed, $$13^2=(-8)^2+15^2+(-8)\cdot15.$$
The second case gives the same values for $b$ and $c$, but $a=-7$. Again, $$13^2=(-7)^2+15^2+(-7)\cdot15.$$
This is a supplement of the answer by @barto.
The problem with the analysis there is that, it is not obvious that $\gcd(2c-d,2c+d)=1.$ So let $m=\gcd(2c-d,2c+d).$ Then we distinguish between two cases:
I. $3\not\mid m$
By the equation $3b^2=(2c-d)(2c+d)$ we observe that $m\mid b$ (actually $m^2\mid b^2,$ hence $(b/m)^2\in\mathbb Z,$ hence $m\mid b$), from which we deduce that $m\mid\gcd(a,b,c)=1,$ thus $m=1.$
II. $3\mid m$
Write $m=3n$ so that $$b^2=3n^2(\frac{2c-d}{m})(\frac{2c+d}{m}).$$ Hence $n^2\mid b^2.$ By the same token as in I. we conclude that $n\mid b.$ Write then $b=b'n$ to find: $$(b')^2=3(\frac{2c-d}{m})(\frac{2c+d}{m}),$$ so $3\mid b'.$ This implies that $m=3n\mid b$ again. So $m\mid1,$ a contradiction.
Therefore, with our assumptions, indeed it cannot happen that $\gcd(2c-d,2c+d)\not=1,$ but this is not so obvious, and is not a mere consequence of the primality of the solution.
Hope this clarifies some point.
