Let $1+a^4+b^4=c^4+d^4$ such that $a$,$b$,$c$ and $d$ are all positive Integers.
How to find Infinitely many solutions ?
By computer search up to $a$,$b$,$c$,$d$ $<3600$ , I found two solutions :
$1+25^4+42^4=17^4+43^4$
$1+405^4+1786^4=253^4+1787^4$
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3$\begingroup$ Another solution is $1+3870^4+11954^4=6953^4+11632^4$. $\endgroup$Peter Mueller– Peter Mueller2025-02-08 20:09:16 +00:00Commented Feb 8, 2025 at 20:09
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2$\begingroup$ Somewhat related are oeis.org/A004831 (Numbers that are the sum of at most 2 nonzero 4th powers. See especially the comment) and oeis.org/A336536 (Numbers n that can be written as both the sum of two nonzero fourth powers and the sum of three nonzero fourth powers.) $\endgroup$Gerry Myerson– Gerry Myerson2025-02-08 21:17:27 +00:00Commented Feb 8, 2025 at 21:17
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2$\begingroup$ There may be something in Andrzej Nowicki, Cubes, Fourth Powers and Higher Powers, available at academia.edu/30783812/Cubes_Fourth_Powers_and_Higher_Powers where 5.2.5. says, Kolejne liczby i sumy dwóch bikwadratów: $3502321 = 25^4 + 42^4$, $3502322 = 17^4 + 43^4$. The Polish translates as Consecutive numbers and sums of two biquadrates. $\endgroup$Gerry Myerson– Gerry Myerson2025-02-08 21:33:51 +00:00Commented Feb 8, 2025 at 21:33
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2$\begingroup$ After Reading a References given by @GerryMyerson in Comments, 1st solution is already exists in oeis and also in Andrzej Nowicki, Cubes, Fourth Powers and Higher Powers. Thanks for References. $\endgroup$Guruprasad– Guruprasad2025-02-09 20:07:02 +00:00Commented Feb 9, 2025 at 20:07
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2$\begingroup$ @David , It works only when factors = 0. Why you think that it has a polynomial parametrization with degree 2 ? $\endgroup$Guruprasad– Guruprasad2025-02-17 09:43:14 +00:00Commented Feb 17, 2025 at 9:43
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1 Answer
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This is not an answer, only some trivial observations: Let $p(n)$ be the number of integers in $\{1,2,\ldots,n\}$ of the form $a^4+b^4$. Then certainly $$ p(n^4)\approx\frac{1}{2}\int_0^n\sqrt[4]{n^4-x^4}dx=\gamma n^2, $$ where $\gamma=\frac{\Gamma(1/4)^2}{16\sqrt{\pi}}=0.4635\ldots$.
Thus $p(n)\approx\gamma\sqrt{n}$. Therefore the expected number of solutions of $1+a^4+b^4=c^4+d^4$ with $a\le b$, $c\le d$ and $a^4+b^4\le n$ should be $$ \approx \sum_{k=1}^n\left(\frac{\gamma}{2\sqrt{n}}\right)^2\approx\frac{\gamma^2}{4}\log n=:E(n). $$ Note that for the third solution we have $n=405^4+1786^4$ and $E(n)=1.608\ldots$, a not so bad approximation of $3$. This may also explain why it is so hard to find a fourth solution. We should have to search up to $n=e^{4\cdot4/\gamma^2}=2.19\ldots\cdot10^{32}$. So $a$ and $b$ should run up to about $10^8$. In fact there is no fourth solution for $a,b\le 1400000$, and given the heuristic, it is not really surprising.
By the way, if this heuristic comes close to the truth, then this would show that there is no polynomial parametrization of solutions.
