Find positive integers $x,y$ and prime $p$ so that $(xy^3)/(x+y)=p$
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1What is the question, and what effort have you made in attempting to answering on your own? (under the assumption that you're asking others to make that effort for you). – barak manos Oct 31 '16 at 14:02
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Excuse me! I tried my best to solve the problem in 1 day, but i couldn't. So i posted it on that website because i need people to help me solve it. – N.Paul Oct 31 '16 at 14:19
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1No problem. Can you please share your attempts, thoughts and observations on the problem? – barak manos Oct 31 '16 at 14:21
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Oh god. I told you i had tried all solutions i thought to solve it but everything was wrong. And i need someone give me an idea. – N.Paul Oct 31 '16 at 14:23
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There is no solution with $1\le x,y\le 10^4$ – Peter Oct 31 '16 at 14:24
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1I don't think that asking for divine intervention is of any use at this point. Please share your observations. For example, what should be the relation between $x$ and $y$ in order for $\frac{(xy)^3}{x+y}$ to be integer? – barak manos Oct 31 '16 at 14:26
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Sorry. I wrote wrongly. But i have fixed it – N.Paul Oct 31 '16 at 14:27
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1I would try to use the fact that for the division to produce an integer (let alone a prime number) all prime factors of $x+y$ must be factors of either $x$ or $y$, and hence both. Do you see why? – Jyrki Lahtonen Oct 31 '16 at 14:30
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1@Peter: OP has changed the original expression. – barak manos Oct 31 '16 at 14:31
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Oh. It's a good idea – N.Paul Oct 31 '16 at 14:36
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It's trickier than I thought initially. $x=14$, $y=2$ yields $p=7$, but are there other solutions? – Jyrki Lahtonen Oct 31 '16 at 14:43
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@JyrkiLahtonen Not, if $1\le x,y\le 10^4$ – Peter Oct 31 '16 at 15:11
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I think that's all. I have just used pascal software to solve it and the result is the same as yours: x=14,y=2 and p=7. But i don't know how to solve it in math – N.Paul Oct 31 '16 at 15:15
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By the way, where did you find this problem? It could be a practice problem for a math contest (in which case it could be tagged [tag:contest-math]). – Jyrki Lahtonen Oct 31 '16 at 16:09
1 Answers
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Let's write $d=\gcd(x,y)$, $x=dx_1$, $y=dy_1$ with $\gcd(x_1,y_1)=1$.
Claim. $y_1=1$.
Proof. Otherwise there exists a prime $\ell\mid y_1$. As $\ell\nmid x_1$ we also have $\ell\nmid (x_1+y_1)$. Let us consider the fraction $$ \frac{xy^3}{x+y}=\frac{d^4x_1y_1^3}{d(x_1+y_1)}=\frac{d^3x_1y_1^3}{x_1+y_1}. $$ Here the denominator is not divisible by $\ell$, but the numerator is divisible by $\ell^3$. Therefore $\ell^3\mid p$ which is absurd. QED.
So $d=y$ and $y\mid x$. Let's write $x=zy$. We get $$ p=\frac{xy^3}{x+y}=\frac{zy^4}{y(z+1)}=\frac{zy^3}{z+1}. $$ For this to be an integer the numerator must be divisible by $z+1$. As $\gcd(z,z+1)=1$ we must have $z+1\mid y^3$. As $p$ is a prime we have two possibilities:
- A) $z=1$ and $y^3=p(z+1)$
- B) $z=p$ and $y^3=z+1$
The case A is impossible, because then $y^3=2p$. In case B we get $p=y^3-1$. Because $y-1\mid (y^3-1)$, $p$ can be a prime only if $y=2$.
This leaves $y=2$, $z=7$, $x=14$, $p=7$ as the only solution.
Jyrki Lahtonen
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@N.Paul Sorry about initially misleading you with that "hint" in a comment. Well, I managed to mislead myself for a while also :-) – Jyrki Lahtonen Oct 31 '16 at 15:58
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I am slightly worried that we might have already had this problem in the form $$p=\frac{zy^3}{z+1}.$$ It somehow rings a bell, but I may be confusing it with another question. Couldn't find it anyway :-/ – Jyrki Lahtonen Oct 31 '16 at 16:05
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That problem is in my midterm test. Anyway, your solution you got is right: x=14, y=2, p=7 – N.Paul Oct 31 '16 at 23:19