Diophantine equation

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#1 h Mar 12, 2025, 1:35 AM

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Find the $5$ smallest positive solutions of $x$ that has an integer $k$ that satisfies $x^2=3k^2+4$

This post has been edited 1 time. Last edited by PaperMath, Mar 12, 2025, 1:36 AM

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#2 h Mar 12, 2025, 1:43 AM

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Using code I found the smallest five solutions:

solutions
x = 4, k = 2
x = 14, k = 8
x = 52, k = 30
x = 194, k = 112
x = 724, k = 418
oops misread the question

This post has been edited 1 time. Last edited by lovematch13, Mar 12, 2025, 1:48 AM

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#3 h Mar 12, 2025, 1:44 AM

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solution

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#4 h Mar 12, 2025, 1:59 AM • 1 Y

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If x,k are integers, it can be solved via Pell equation $\left(\frac{x}{2}\right)^2-3 \left(\frac{k}{2}\right)^2=1$

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#5 h Mar 12, 2025, 3:27 AM

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Pls stop spamming trivial questions... just plug in values of k

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#6 h Mar 12, 2025, 1:06 PM

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derekli wrote:

Pls stop spamming trivial questions... just plug in values of k

Ok then show us how you plug every value of k

This post has been edited 1 time. Last edited by PaperMath, Mar 12, 2025, 1:06 PM

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#7 h Mar 12, 2025, 1:11 PM

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derekli wrote:

Pls stop spamming trivial questions... just plug in values of k

Yeah just stop. It's not trivial for some people and not everyone is as good at math as you are.

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#8 h Mar 12, 2025, 1:20 PM

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Held me with ä forom

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jlcong

#9 h Yesterday at 7:01 PM

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PaperMath wrote:

derekli wrote:

Pls stop spamming trivial questions... just plug in values of k

Ok then show us how you plug every value of k

Rearrange and plug in k=0, 1, 2, 3, 4, 5. These create the smallest positive values of x! :-D

:-D

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#10 h Yesterday at 7:25 PM

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$x^2=3k^2+4$ $x^2-3k^2=4$ $(x+\sqrt{3}k)(x-\sqrt{3}k)=4$ The cases-
1) $x+\sqrt{3}k = 4, x-\sqrt{3}k =1 \implies x$ isn't an integer
2) $x+\sqrt{3}k = 1, x-\sqrt{3}k=4 \implies x$ isn't an integer
3) $x+\sqrt{3}k=2, x-\sqrt{3}k=2 \implies x=2$
4) $x+\sqrt{3}k=-4, x-\sqrt{3}k =-1 \implies x$ isn't an integer
5) $x+\sqrt{3}k = -2, x-\sqrt{3}k=-2 \implies x=-2$ which doesn't satisfy
(Assuming $x$ is a positive integer and $k$ is some integer)
Without $x$ being a positive integer, it's possible to get $x=\frac {5} {2}$ . This is an incomplete solution I think.

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