Find all pairs of co-primes $x,y$ such that $\frac{5y^2}{x^2-xy}$ is a square

Question

Find all pairs of natural coprime numbers for which $\sqrt{\frac{5y^2}{x^2-xy}}$ is a natural number.

My approach is that for it to be natural $x^2-xy \mid 5y^2$ so then $x$ is either $1$ or $5$. Then when $x=1 , y$ is either $0,-4,2$ or $6$. I'm not sure which one of these values I should choose.

Your condition is not sufficient. You must have that $5y^2 = x(x-y)z^2$ for some natural number $z$. — PierreCarre, Dec 30 '23 at 14:03
@Student Only if $z$ is the square of something, this is why I proposed $z^2$ right from the start. — PierreCarre, Dec 30 '23 at 14:10
I added an answer that should make the proof "obvious" (and highlights the key role played by coprimality and Euclid's Lemma). — Bill Dubuque, Dec 30 '23 at 22:19

score 1 · Accepted Answer · answered Dec 30 '23 at 14:23

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We must have that $5y^2 = x(x-y)z^2$ for some natural number $z$. Since $x,y$ are co-primes, $x$ must be either 1 or 5. Since $x=1$ does not work (substitute and see), we conclude that $x$ must be 5. Looking at the resulting condition, $y^2 = (5-y) z^2$, we just need to try values of $y$ smaller than 5.

answered Dec 30 '23 at 14:23

PierreCarre

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Oops sorry I missed that (+1) – Anne Bauval Dec 30 '23 at 16:10
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Note: as in the question, the argument implicitly uses Euclid's Lemma (or equivalent), i.e. $(x,y)=1\Rightarrow (x,y^2)=1,$ so $,x\mid 5y^2\Rightarrow x\mid 5\ \ $ – Bill Dubuque Dec 30 '23 at 19:02
I added an answer which highlights the key role of $\rm\color{#c00}{Euclid's\ Lemma} \ $ – Bill Dubuque Dec 30 '23 at 20:39

Bill Dubuque · Answer 2 · 2023-12-30T21:58:55.373

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Note $\,x(x\!-\!y)\mid py^2$ so $\,x\!=\!p,\,x\!-\!y\!=\!1\,$ by $(x,y)\!=\!1$ and below (OP has $\,p\!=\!5)$.

Theorem $\,ab\mid pc\,\Rightarrow\, a\!=\!p,\, b\!=\!1\,$ if $\,a\!>\!b\!>\!0\,$ are coprime to $\,c\,$ and $\,p\,$ is prime.

Proof $ $ By coprimality and Euclid's Lemma $\,a,b\mid p\,$ so $\,a\!=\!p,\,b\!=\!1\,$ by $\,a\!>\!b\!>\!0$.

edited Dec 30 '23 at 21:58

answered Dec 30 '23 at 20:38

Bill Dubuque

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Therefore the quotient $,pc/ab = c,,$ so it is a square simply because $c = y^2$ is a square in the OP. $ $ Notice $,(x,y)!=!1,\Rightarrow,x,,x!−!y,$ are coprime to $,y,$ so coprime to $,y^2,$ again by Euclid (or by gcd laws). $\ \ $ – Bill Dubuque Dec 30 '23 at 22:26
Thank you . so y has to be a prime number smaller than 5 right? – Student Jan 02 '24 at 10:55
@Student OP has $p=5$ so by the Theorem we get $x=5,,x-y=1\Rightarrow y = 4.,$ You don't need to search for $y$ as in the other answer. – Bill Dubuque Jan 02 '24 at 11:51
Thank you so much! – Student Jan 02 '24 at 15:45

Find all pairs of co-primes $x,y$ such that $\frac{5y^2}{x^2-xy}$ is a square

2 Answers2