# Parametric Diophantine solutions of $kx^4+1=y^2$.

I’ve seen a number of question asking for solutions of $kx^4+1=y^2$ where $k$ is a small fixed value, and found these frustrating when I find just two, or fewer solutions.

Just for fun, I wondered what happened when I fixed $x$ to investigate what values of $k$ were produced. There are far more than I presumed, and (after rather more effort than I expected), some fit into parametric solutions.

In the following $a$ and $b>2$ are positive integers and the pairs of $\pm$ are both plus or both minus.

$$(k,x,y)=(a^2-1,1,a)$$

$$(k,x,y)=(a(2^4a\pm2),2,2^4a\pm1)$$

$$(k,x,y)=(a(2^6a\pm1),4,2^7a\pm1)$$

$$(k,x,y)=(a(2^{4b-2}a\pm1),2^b,2^{4b-1}a\pm1)$$

Examples for $(k,x,y)=(a^2-1,1,a)$ with $a=2,3,4,5,6,7$

$$(3,1,2)$$ $$(8,1,3)$$ $$(15,1,4)$$ $$(24,1,5)$$ $$(35,1,6)$$ $$(48,1,7)$$

Examples for $(k,x,y)=(a(2^4a\pm2),2,2^4a\pm1)$ with $a=1,2,3$, minus then plus.

$$(14,2,15)$$ $$(60,2,31)$$ $$(138,2,47)$$ $$(18,2,17)$$ $$(68,2,33)$$ $$(150,2,49)$$

Examples for $(k,x,y)=(a(2^6a\pm1),4,2^5a\pm1)$ with $a=1,2,3$, minus then plus.

$$(63,4,127)$$ $$(254,4,255)$$ $$(573,4,383)$$ $$(65,4,129)$$ $$(258,4,257)$$ $$(579,4,385)$$

Examples for $(k,x,y)=(a(2^{4b-2}a\pm1),2^b,2^{4b-1}a\pm1)$ with $b=3,a=1,2,3$, minus then plus.

$$(1023,8,2047)$$ $$(4094,8,4095)$$ $$(9213,8,6143)$$ $$(1025,8,2049)$$ $$(4098,8,4097)$$ $$(9219,8,6145)$$

Examples for $(k,x,y)=(a(2^{4b-2}a\pm1),2^b,2^{4b-1}a\pm1)$ with $b=4,a=1,2,3$, minus then plus.

$$(16383,16,32767)$$ $$(65534,16,65535)$$ $$(147453,16,98303)$$ $$(16385,16,32769)$$ $$(65538,16,65537)$$ $$(147459,16,98305)$$

Examples for $(k,x,y)=(a(2^{4b-2}a\pm1),2^b,2^{4b-1}a\pm1)$ with $b=5,a=1,2,3$, minus then plus.

$$(262143,32,524287)$$ $$(1048574,32,1048575)$$ $$(2359293,32,1572863)$$ $$(262145,32,524289)$$ $$(1048578,32,1048577)$$ $$(2359299,32,1572865)$$

Thanks to a comment from @JyrkiLahtonen I’ve found, and corrected, a program bug that incorrectly gave solutions for some high values. Hence, I’ve removed the restriction that $k$ not be square, as it’s unnecessary.

My question:

Please could you find more parametric solutions to $kx^4+1=y^2$ ? I can almost detect patterns, but just can’t resolve them.

• This is undoubtedly interesting, but what was the question again? For a fixed $x$ you always get infinitely many solutions $(k,y)$, because whenever $y\equiv\pm1\pmod{x^4}$ we get a solution for some $k$. For a fixed $k$ we get only finitely many solutions, because then the curve is elliptic (not sure about the field of definition), and those can only have finitely many integer points. It is rare for $k$ to be a square, because then both $kx^4$ and $y^2$ are squares, and squares differ from each other by one only when they are $0$ and $1$. – Jyrki Lahtonen Jun 5 '17 at 20:40
• @JyrkiLahtonen Well spotted, thank you. I’ve updated my question to add my reason for my restriction on $k$, but will remove it if people think it advisable. – Old Peter Jun 6 '17 at 9:31
• Maybe pell equation would give a direction? – Mudream Jun 6 '17 at 10:04
• @Mudream Thank you so much for your interest. I can't see how to use the Pell equation, but I have no experience in using it, so maybe I'm missing something. – Old Peter Jun 7 '17 at 12:54
• I’ve found another parametric solution. Should I add it to the question, or as an answer? – Old Peter Jun 7 '17 at 12:57

This is only an attempt at a partial answer.

I’ve found a solution that gives, IMHO, an infinite set of results for every positive value of $x$.

$$(k,x,y)=(a^2x^4\pm2a,x,ax^4\pm1)$$

where $a$ is a positive integers and the pairs of $\pm$ are both plus or both minus, except for $x=1$ where both are plus only.

However, this does not cover all solutions, except perhaps for $x=1$.

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I’ve also found solutions for individual values of $x$, but feel these are of limited value or interest.

For example, with $x=2$, we have, $$(k,x,y)=( 16a^2-18a+5,2,16a-9)$$ $$(k,x,y)=( 16a^2-14a+3,2,16a-7)$$

Taken with $$(k,x,y)=(a(2^4a\pm2),2,2^4a\pm1)$$

these cover all the solutions for $x=2$ that I’ve found.