WakeUp wrote: Let $m$ be a positive integer such that $m=2\pmod{4}$. Show that there exists at most one factorization $m=ab$ where $a$ and $b$ are positive integers satisfying \[0<a-b<\sqrt{5+4\sqrt{4m+1}}\] we have $0<a-b< \sqrt{5+4\sqrt{4m+1}}$ so $0< (a-b)^2< 5+4\sqrt{4m+1}$ so $4m< (a+b)^2< 5+4m+4\sqrt{4m+1}=(\sqrt{4m+1}+2)^2$ so $\sqrt{4m}< (a+b)< \sqrt{4m+1}+2$ but 4m is no a square so $[\sqrt{4m}]+1\le (a+b)\le [\sqrt{4m+1}]+2$ let $[\sqrt{4m}]=a$ then $a<\sqrt{4m}<a+1$ then $a^2<4m<a^2+2a+1$ then $a^2+1\le 4m\le a^2+2a$ then $a^2+2\le 4m+1\le a^2+2a+1$ then $[\sqrt{4m+1}]=a, (a+1)$ from here we will see that $[\sqrt{4m+1}=a+1$ when $4m=a^2+2a$ or $m=t(t+1)$ assume that there existed: (a,b) and (c,d) which satisfied the condition ** Case 1: $[\sqrt{4m+1}]=a$ then $a+b=c+d+1$ but $m\equiv 2 (mod 4)$ then a+b,c+d is odd ,contradiction ** Case 2 $[\sqrt{4m+1}]=a+1$ then $m=t(t+1)$ then $2t+1\le a+b<2t+3$ so it just like case 1 which is wrong *** q.e.d

WakeUp wrote: Let $m$ be a positive integer such that $m=2\pmod{4}$. Show that there exists at most one factorization $m=ab$ where $a$ and $b$ are positive integers satisfying \[0<a-b<\sqrt{5+4\sqrt{4m+1}}\] This is wrong for $m=2p$ where $p$ is an odd large prime.

nnosipov wrote: WakeUp wrote: Let $m$ be a positive integer such that $m=2\pmod{4}$. Show that there exists at most one factorization $m=ab$ where $a$ and $b$ are positive integers satisfying \[0<a-b<\sqrt{5+4\sqrt{4m+1}}\] This is wrong for $m=2p$ where $p$ is an odd large prime. are u sure? the question is "at most one " not exist one.

tuandokim wrote: nnosipov wrote: WakeUp wrote: Let $m$ be a positive integer such that $m=2\pmod{4}$. Show that there exists at most one factorization $m=ab$ where $a$ and $b$ are positive integers satisfying \[0<a-b<\sqrt{5+4\sqrt{4m+1}}\] This is wrong for $m=2p$ where $p$ is an odd large prime. are u sure? the question is "at most one " not exist one. Sorry, I agree.