Let $a,b$ be two different positive integers. Suppose that $a,b$ are relatively prime. Prove that $\dfrac{2a(a^2+b^2)}{a^2-b^2}$ is not an integer.
Problem
Source: 2019 Thailand Mathematical Olympiad P2
Tags: number theory
MarkBcc168
22.05.2019 13:56
The main claim is
Claim: (Gcd bash) We have $\gcd(a^2-b^2, a(a^2+b^2))\mid 2$.
Proof: Let $d\mid a^2-b^2$ and $d\mid a(a^2+b^2)$. We get
\begin{align*}
d\mid a(a^2+b^2) + a(a^2-b^2) &\implies d\mid 2ab^2 \\
d\mid 2a(a^2+b^2) &\implies d\mid 2a^3 \\
d\mid a^2-b^2\mid 2(a^4-b^4) &\implies d\mid 2b^4
\end{align*}thus $d\mid 2\gcd(a,b)^4=2$ so we are done.
By the claim, we get $a^2-b^2\mid 4$ which is a clear contradiction.
ytChen
28.05.2019 04:19
MarkBcc168 wrote: Let $a,b$ be two different positive integers. Suppose that $a,b$ are relatively prime. Prove that $\dfrac{2a(a^2+b^2)}{a^2-b^2}$ is not an integer. Solution. To seek a contradiction, assume the contrary $a^2-b^2\mid2a(a^2+b^2)$. The given condition $\gcd(a,b)=1$ gives $$\gcd(a^2-b^2,a)=1=\gcd(a^2-b^2,a^2),$$from which it follows that \begin{align*}a^2-b^2\mid2a(a^2+b^2)&\implies a^2-b^2\mid2(a^2+b^2) \\ &\implies a^2-b^2\mid2(a^2+b^2)+2(a^2-b^2)=4a^2 \\ &\implies a^2-b^2\mid 4\\ &\implies a+b\mid 4. \end{align*}Since that $a,b$ are different positive integers, we have $a+b\ge3$. Therefore, $a+b\mid 4$ yields $a+b=4$, which leads to $a-b=\pm1$ by the result of $a^2-b^2\mid 4$. Thus we get $2a=(a+b)+(a-b)=4\pm1$. But even number $\ne$ odd number, a contradiction. $\blacksquare$
Pluto1708
03.07.2019 15:32
Pls veriy this
Assume the contrary.Let $p|a^2-b^2$ be a prime.The case when $a^2-b^2$ is a prime is easy to deal with. otherwise Clearly $p$ doesn't divide $a$ or $b$..Also its easy to see that $p|a-b$ or $p|a+b$ not both.WLOG let $p|a-b$.Then $$\dfrac{a((a+b)^2+(a-b)^2))}{a^2-b^2} \implies \dfrac{a((a+b)^2+(a-b)^2))}{p} \implies \dfrac{a(a+b)^2}{p} \in \mathbf{Z} \implies p|(a+b)$$a contradiction so we are done.$\blacksquare$
Thanks @below!
XbenX
03.07.2019 15:58
Pluto1708 wrote: Pls veriy this
Assume the contrary.Let $p|a^2-b^2$ be a prime.The case when $a^2-b^2$ is a prime is easy to deal with. otherwise Clearly $p$ doesn't divide $a$ or $b$..Also its easy to see that $p|a-b$ or $p|a+b$ not both.WLOG let $p|a-b$.Then $$\dfrac{a((a+b)^2+(a-b)^2))}{a^2-b^2} \implies \dfrac{a((a+b)^2+(a-b)^2))}{p} \implies \dfrac{a(a+b)^2}{p} \in \mathbf{Z} \implies p|(a+b)$$a contradiction so we are done.$\blacksquare$
You are considering the case when $p$ is odd, otherwise we would not always have $\gcd(a-b,a+b)=1$, but this case is easy to deal with. Assume all prime divisors of $a^2-b^2$ are even then $a,b$ are odd and $a^2-b^2 \mid 4ab^2$ wich clearly has no solutions.
ubermensch
03.07.2019 16:01
Hmm... seems like quite a simple problem... First, you get $gcd(a^2-b^2, 2a^3+2a^2b)=(a^2-b^2,4a^2b)$. Now, as $(a^2-b^2,a^2)=(b^2,a^2)=1$ and $(a^2-b^2,b)=(a^2,b)=1$, hence we get that $a^2-b^2$ divides $4$, which is obviously false as $a,b>0$.
Rizsgtp
06.07.2019 12:03
raghoodah1m
14.01.2020 06:33
ytChen wrote: MarkBcc168 wrote: Let $a,b$ be two different positive integers. Suppose that $a,b$ are relatively prime. Prove that $\dfrac{2a(a^2+b^2)}{a^2-b^2}$ is not an integer. Solution. To seek a contradiction, assume the contrary $a^2-b^2\mid2a(a^2+b^2)$. The given condition $\gcd(a,b)=1$ gives $$\gcd(a^2-b^2,a)=1=\gcd(a^2-b^2,a^2),$$from which it follows that \begin{align*}a^2-b^2\mid2a(a^2+b^2)&\implies a^2-b^2\mid2(a^2+b^2) \\ &\implies a^2-b^2\mid2(a^2+b^2)+2(a^2-b^2)=4a^2 \\ &\implies a^2-b^2\mid 4\\ &\implies a+b\mid 4. \end{align*}Since that $a,b$ are different positive integers, we have $a+b\ge3$. Therefore, $a+b\mid 4$ yields $a+b=4$, which leads to $a-b=\pm1$ by the result of $a^2-b^2\mid 4$. Thus we get $2a=(a+b)+(a-b)=4\pm1$. But even number $\ne$ odd number, a contradiction. $\blacksquare$ why is that it $a+b=4$ ? it can be any number Divide by 4