Beautiful number. $\frac{x^2+y^2}{x+y}=a$ Solutions have the form: $a=p^2+s^2$ $y=s(p+s)$ $x=p(p+s)$

individ wrote: Beautiful number. $\frac{x^2+y^2}{x+y}=a$ Solutions have the form: $a=p^2+s^2$ $y=s(p+s)$ $x=p(p+s)$ False... $15$ ($x=18$ , $y=6$) is a beautifull number and not represanted as sum of two squares because $3$ is prime and $3=3mod4$ and $U_3(15)=1$ Lets see... $x^2+y^2=a(x+y)$ wlog $x>y$ So $(x+y)^2-2a(x+y)+(x-y)^2=0$ $D=4a^2-4(x-y)^2=4w^2$ $a^2=(x-y)^2+w^2$ With $x-y>0$ So there are positive integers $u,v$ not the same parity and $u>v$ and $(u,v)=1$ and $d$ positive integer such as $a=d(x^2+y^2)$ $x-y=2duv$ $w=d(u^2+v^2)$ Or $a=d(u^2+v^2)$ $x-y=d(u^2-v^2)$ $w=2duv$ From 2 systems finally we get two solutions...For $x>y$ $(a,x,y)=(d(u^2+v^2),du(u+v),du(u-v))$ $(a,x,y)=(d(u^2+v^2),du(u+v),dv(u+v))$ Where $u,v$ naturals with $(u,v)=1$ not the same parity ,$u>v$ and $d$ positive integer.

chaotic_iak wrote: A positive integer is called beautiful if it can be represented in the form $\dfrac{x^2+y^2}{x+y}$ for two distinct positive integers $x,y$. A positive integer that is not beautiful is ugly. a) Prove that $2014$ is a product of a beautiful number and an ugly number. b) Prove that the product of two ugly numbers is also ugly. a) we must solve the equation $d(a^2+b^2)=2014=2*19*53$ So $d=1,2,19,2014$ no solutions. b) supose that $n$ is a beautifull and $n=d(a^2+b^2)=mn$ for some ugly $m,n$ So $d|mn$ and $a^2+b^2=\dfrac{mn}{d}$ with $(a,b)=1$ It is known that all factors of $a^2+b^2$ with $(a,b)=1$ are sum of two squares ...but ugly number not. Contradiction.

What is it? What needs to be explained? Another solution can be obtained by multiplying and dividing all the numbers by the same amount. Αρχιμήδης 6 wrote: $(a,x,y)=(d(u^2+v^2),du(u+v),dv(u+v))$ This is the formula which I brought. $a=p^2+s^2$ $y=s(p+s)$ $x=p(p+s)$ When replacing $p\longrightarrow{p+s}$ : $s\longrightarrow{p-s}$ and a reduction of 2. $a=p^2+s^2$ $y=p(p-s)$ $x=p(p+s)$ It turns out the same: Αρχιμήδης 6 wrote: $(a,x,y)=(d(u^2+v^2),du(u+v),du(u-v))$ Comment is funny: Αρχιμήδης 6 wrote: False... $15$ ($x=18$ , $y=6$) is a beautifull number and not represanted as sum of two squares because If $p=3$ : $s=1$ $a=9+1=2*5$ $y=1(3+1)=2*2$ $x=3(3+1)=2*6$ This is: $a=15=3*5=3(2^2+1^2)$ $x=3*6=3*2*(2+1)$ $y=3*2=3*2*(2-1)$ Before you scold - check everything!

individ wrote: Beautiful number. $\frac{x^2+y^2}{x+y}=a$ Solutions have the form: $a=p^2+s^2$ $y=s(p+s)$ $x=p(p+s)$ Is so difficult to understand that $15$ is beautifull and cant represent as sum of two squares? At your post you said that $a$ is sum of two squares and it is FALSE!!! If you believe that $15$ is sum of two squares then you don't work in integers. Or you believe that $d(a^2+b^2)$ is equal to $a^2+b^2$? If you can't understand this simle statement i can't help you any more...

When I write for the equation: $a^2+b^2=c^2$ Write the solution in this form: $a=p^2-s^2$ $b=2ps$ $c=p^2+s^2$ Everyone understands that you can write not mutually simple solutions: $a=d(p^2-s^2)$ $b=d2ps$ $c=d(p^2+s^2)$ What we will say с=15 cannot be represented as a sum of two squares? And with the high number of Pythagorean triples not? If the formula is written, it is written with the statement that decisions are coprime. Not coprime get trivial. Simply by multiplication.

individ's logic

Let us first completely identify the beautiful numbers. Take $d=\gcd(x,y)$, $x=da$, $y=db$, with $\gcd(a,b) = 1$ and moreover $(a,b) \neq (1,1)$ since we want $x\neq y$. Since we need $x+y \mid x^2+y^2$, that writes as $a+b \mid d(a^2+b^2)$. $\bullet$ If $a,b$ are of different parity then $\gcd(a+b,a^2+b^2) = 1$, so then we need $a+b \mid d$, thus $d=e(a+b)$. Thus $x=e(a+b)a$, $y=e(a+b)b$, and $\dfrac {x^2+y^2}{x+y} = e(a^2+b^2)$. $\bullet$ If $a,b$ are both odd then $\gcd(a+b,a^2+b^2) = 2$, so then we need $\dfrac {a+b}{2} \mid d$, thus $d=\dfrac {1}{2}e(a+b)$. Thus $x=\dfrac {1}{2}e(a+b)a$, $y=\dfrac {1}{2}e(a+b)b$, and $\dfrac {x^2+y^2}{x+y} = \dfrac {1}{2}e(a^2+b^2) = e\left (\left (\dfrac {a+b}{2}\right )^2 + \left (\dfrac {a-b}{2}\right )^2\right )$, of the same type $\dfrac {x^2+y^2}{x+y} = e(a'^2+b'^2)$ as above, with $a'= \dfrac {a+b}{2}$, $b'= \dfrac {|a-b|}{2}$, and $a',b'$ of different parities. This shows that the beautiful numbers $n$ are those and only those that can be represented as $n = e(a^2+b^2)$ for $a,b$ coprime and of different parity (thus distinct). A direct consequence is that the product of a beautiful number $n = e(a^2+b^2)$ with any other number $m$ is beautiful, since $mn = (me)(a^2+b^2)$. Obviously $1$ is ugly and $2$ is ugly. An odd prime $p\equiv 1\pmod{4}$ is beautiful, since there (uniquely) exist $a,b$ (coprime and of different parity) such that $p=a^2+b^2$ and we can take $e=1$. Finally, an odd prime $p\equiv -1\pmod{4}$ is ugly. We can now push the characterization even further. If a number $n$ has a prime factor $p\equiv 1\pmod{4}$ then it is beautiful. Conversely, a beautiful number $n=e(a^2+b^2)$ has such a prime factor of $a^2+b^2$. Therefore the beautiful numbers are precisely those that have a prime factor $p\equiv 1\pmod{4}$. Thus the point b) of the problem is solved, and the point a) follows since $2014 = 2\cdot 19\cdot 53$; thus $53, 2\cdot 53, 19\cdot 53, 2\cdot 19\cdot 53$ are beautiful, while $2\cdot 19, 19, 2, 1$ are ugly, and we have multiple choices for the required representation.

chaotic_iak wrote: A positive integer is called beautiful if it can be represented in the form $\dfrac{x^2+y^2}{x+y}$ for two distinct positive integers $x,y$. A positive integer that is not beautiful is ugly. a) Prove that $2014$ is a product of a beautiful number and an ugly number. b) Prove that the product of two ugly numbers is also ugly. $\\x=k(1-p)$ $\\y=k(1+p)$