Are there integers $a,c,d,e,f$ such that $$a^2 + \left(\dfrac{c}{d}\right)^2= \left(\dfrac{e}{f}\right)^2,\quad\gcd(c,d)=1=\gcd(e,f),\quad f\neq 1\neq d, ~~~ f\neq d $$ As for a contradiction we would have $$a^2 = \left(\dfrac{e}{f}-\dfrac{c}{d}\right) \left(\dfrac{e}{f}+\dfrac{c}{d}\right) = \left(\dfrac{ed-cf}{fd}\right)\left(\dfrac{ed+cf}{fd}\right).$$ Thus $(fd)^2\mid\biggl((ed-cf)(ed+cf)\biggr)$ which happens if $fd\mid (ed-cf)$ or $fd\mid (ed+cf)$. We know $fd$ cannot divide $ed$ because $$fd\cdot k = ed \Rightarrow f \cdot k = e\Rightarrow k = \dfrac{e}{f} \Rightarrow \gcd(e,f)>1$$ But if $fd$ cant divide $ed$ does that mean $fd$ cannot divide the sum or difference $ed+cf$ or $ed-cf$. Instantly, we should say $a^2$ is not an integer which is a contradiction.
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Please don't try to micromanage spacing in title lines. The website uses responsive design to help it be readable across multiple screen formats, but micromanaging spacing within formulas by adding unbreakable spaces hobbles it. – Arturo Magidin Dec 23 '23 at 19:29
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@ArturoMagidin sorry about that – Roy Burson Dec 23 '23 at 19:31
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@peterwhy yes sorry that is not allowed – Roy Burson Dec 23 '23 at 19:32
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1$1^2+\left(\frac{5}{12}\right)^2=\left(\frac{13}{12}\right)^2$. – Anurag A Dec 23 '23 at 19:42
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2Do you allow $f=d \ne 1$? Maybe $$1^2 + \left(\frac{4}{3}\right)^2 = \left(\frac{5}{3}\right)^2$$ – GEdgar Dec 23 '23 at 19:42
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1In your reasoning what if $ed+cf$ is a multiple of $f^2d^2$ for instance ? You separated the two terms as if they needed to be both integers, but it is possible that one rational factor compensates for the other. – zwim Dec 23 '23 at 19:46
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@GEdgar I think about restricting $a\neq 1$ because im searching for semi congruent numbers and as a consequence the number $n= 1/2 a \dfrac{c}{d}$ would be congruent number so we need $d\vert a$ or $a\neq 1$ – Roy Burson Dec 23 '23 at 19:47
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1For $a\ne 1$, maybe $$2^2 + \left(\frac{3}{2}\right)^2 = \left(\frac{5}{2}\right)^2$$ – peterwhy Dec 23 '23 at 19:49
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1Restriction $a \ne 1$ by itself does not help. $$5^2 + \left(\frac{20}{3}\right)^2 = \left(\frac{25}{3}\right)^2$$ – GEdgar Dec 23 '23 at 19:50
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@peterwhy what about $f\neq d$ . Must we have $f=d$ for any solution? – Roy Burson Dec 23 '23 at 19:53
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@GEdgar what about $f\neq d$? – Roy Burson Dec 23 '23 at 19:58
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@zwim oh I see yea that was mistake I guess – Roy Burson Dec 23 '23 at 20:01
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No, $f=d$ must hold. – GEdgar Dec 23 '23 at 20:21
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@GEdgar Why though here do we need $fd \vert cf $? – Roy Burson Dec 23 '23 at 20:28
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1By here in the linked dupe, since the reduced fractions $,e^2/f^2, \ -c^2/d^2,$ sum to an integer they have equal denominators. – Bill Dubuque Dec 23 '23 at 22:48
2 Answers
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Not possible. Suppose $f \ne d$ and $f \ne -d$. Say $f$ does not divide $d$. From $$ a^2 + \left(\dfrac{c}{d}\right)^2= \left(\dfrac{e}{f}\right)^2 $$ we get $$ (ad)^2 + c^2= \left(\dfrac{ed}{f}\right)^2 $$ an integer equal to a non-integer. Similarly for the case $d$ does not divide $f$.
GEdgar
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Correct, you did not, and neither did I. Is this not right: if $gcd(f,e)=1$ and $f|ed$, then $f|d$ ? – GEdgar Dec 23 '23 at 20:36
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the logic doesn't seem right we just found that f must divide d we didnt show there is not a solution – Roy Burson Dec 23 '23 at 20:46
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1@Life_is_pie As you stated, since the answer shows $f\mid d$, then $d = fk$ for some integer $k$. The answer also proves that $d\mid f$, so $f=dm$ for some integer $m$. Combining the $2$ equations gives, since $d\neq 0$, that $d = (dm)k = dmk ;\to; 1 = mk$. Thus, $m = k = \pm 1$, meaning we need to have $f=\pm d$. – John Omielan Dec 23 '23 at 21:11
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@JohnOmielan okay that makes since. You can multiply by $d^2$ or $f^2$ – Roy Burson Dec 23 '23 at 21:17
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@GEdgar I will accept you answer because you were aware of both equations even though you wrote one and the solution is derived from your idea. – Roy Burson Dec 23 '23 at 21:36
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Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Dec 23 '23 at 22:50
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Ignoring the trivial equations $0^2 + 0^2 = 0^2$ , $y^2 + 0 = y^2$ and $0^2 + x^2 = x^2$, we have that $a,c,d,e,f \neq 0$.
So we want to solve
$$ a^2 + (c/d)^2 = (e/f)^2 $$
by multiplying both sides by $d^2$ or by $f^2$ we arrive at the two equations :
$$ (ad)^2 + c^2 = (ed/f)^2 $$
and
$$(af)^2 + (cf/d)^2 = e^2$$
We must conclude that since $\gcd(e,f) = 1$ in the first equation and the LHS is an integer , we get that $f|d$. Likewise for the second equation by $\gcd(c,d)=1$ we get that $d|f$.
However if $f|d$ and $d|f$ and $f,d$ are nonzero, we get either $f=g$ or a contradiction.
But you wanted $f \neq d$ so your equation has no nontrivial solution.
mick
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your proof is more accurate (complete) but it was derived from GEdgar answer. – Roy Burson Dec 23 '23 at 21:38
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@Life_is_pie I did not even read his answer until 5 min ago. I do not need help with trivial number theory. Avoiding divisibility in easy number theory is almost impossible, unless you want an unnessary long or complicated proof. I guess I am a bit offended by " derived ' and you just wanted a reason/motivation to accept his answer instead of mine. That's ok, you can accept his answer. He was first and correct, so its ok. – mick Dec 23 '23 at 22:02
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I would also like to modify question and look for solutions when $d\vert a$. I will ask this is next question because as a consequence the number $n = 1/2 a \dfrac{c}{d}$ would be congruent. For this to happen we need to look for solutions when $d\vert a$. – Roy Burson Dec 23 '23 at 22:04
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As for your variant of the question, that can be resolved in a similar way. – mick Dec 23 '23 at 22:07
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I would like to add that when I answered you did not accept any answers, so I felt a more clear answer was needed. – mick Dec 23 '23 at 22:09
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at the time I was still gathering my thoughts on why his is correct. It's because he said "similarly" which indicates he was aware of both equations even though he didn't write it. Which would lead us to the same conclusion – Roy Burson Dec 23 '23 at 22:13
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1I do as well as I said yours is more accurate but It is more appropriate to give him the points. I just posted next question here https://math.stackexchange.com/questions/4832727/are-there-any-semi-congruent-numbers – Roy Burson Dec 23 '23 at 22:25
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Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Dec 23 '23 at 22:50