I have been struggling on this homework question for quite a while.
Find $x,y$ if LCM $= x+y$ and HCF $= x-y$
Here is my working:
Since, LCM $\times$ HCF $= xy$,
$$ (x+y)(x-y) = xy \\ x^2 - y^2 = xy $$
I am stuck here...
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1@BrianMoehring My school teacher gave me this question, I also thought it's impossible. – athrb Sep 20 '21 at 20:55
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Hint: $x^2-y^2=xy$ is equivalent to $\left(\dfrac xy\right)^2-1=\dfrac xy$. – Sep 20 '21 at 21:19
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@IAmAGuest True, if $y\ne 0$. $x=y=0$ is a perfectly valid solution for that equation. – Andrei Sep 20 '21 at 21:40
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@Andrei: goes without saying. – Sep 21 '21 at 06:41
1 Answers
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Let's call $z=HCF=\gcd(x,y)$. Then $$x=x_1z\\y=y_1z$$where $\gcd(x_1,y_1)=1$. Then we have $$HCF=x-y\\z=x_1z-y_1z$$ Assuming $z\ne 0$, then $$x_1-y_1=1$$ Proceed in the same way with the other information:$$LCM=x+y\\x_1y_1z=x_1z+y_1z$$ Same as before, if $z\ne 0$, you get $$x_1y_1=x_1+y_1$$ Using $x_1=1+y_1$ from a previous equation, we get $$y_1(1+y_1)=1+y_1+y_1$$or $$y_1^2-y_1-1=0$$ This equation does not have integer solutions. So we need to go back and check the $z=0$ case, where $x=y=0$.
Andrei
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