# A curious coincidence for Wroblewski's solutions to $1^4+x_2^4+x_3^4+x_4^4+x_5^4 = y_1^4$

J. Wroblewski has a database for the equation, $$x_1^4+x_2^4+x_3^4+x_4^4+x_5^4 = y_1^4$$ Of about $13700$ primitive solutions with $y_1 < 10000$ and $y_1\not\equiv 0{\pmod 5}$, only four examples can be found with $x_1 = 1$, namely $$1^4 + 150^4 + 3340^4 + 6130^4 + 6350^4 = 7499^4\\ 1^4 + 2520^4 + 3250^4 + 5050^4 + 6970^4 = 7499^4\tag1$$ and $$1^4 + 920^4 + 3120^4 + 5410^4 + 8870^4 = 9193^4\\ 1^4 + 1410^4 + 3490^4 + 6020^4 + 8680^4 = 9193^4\tag2$$ Surely the fact they come in pairs can't be coincidence. I think they are particular instances of a identity similar to cubes, $$\small 1^3 + (-1 - 9t^3 - 648t^6 + 3888t^9)^3 + (3t + 81t^4 - 1296t^7 + 3888t^{10})^3 = (-135t^\color{red}4 + 3888t^\color{red}{10})^3\tag3$$ Because the RHS has only $\color{red}{even}$ powers, then it is immune to sign changes. For example, let $t=-1,1$ and we get, $$1^3 -4528^3 + 5262^3 = 3753^3\\ 1^3 + 3230^3 + 2676^3 = 3753^3$$ So that explains the one for cubes.

Q: What, however, explains $(1)$ and $(2)$? And if an identity is behind it, then what is it?

P.S. If $y_1\equiv 0\pmod5$, there are more solutions, the smallest again comes in a pair, $$1^4 + 8^4 + 24^4 + 36^4 + 38^4 = 45^4\\ 1 ^4 + 2^4 + 12^4 + 24^4 + 44^4 = 45^4$$ with the first one missed in Mathworld's list.

• Could the two pairs perhaps result from an identity involving quadratic forms? RHS = $(ax^2+bxy+cy^2)^4$ with $a+c=8346$, $b=847$ and $(x,y)=(\pm1,\pm1)$ would give each of $7499^4$ and $9193^4$ twice. – Adam Bailey Dec 16 '16 at 16:15
• @AdamBailey: That's a very interesting observation. I'll think about it. – Tito Piezas III Dec 16 '16 at 16:37

I assume primitive solutions, and use $y$ for $y_1$, throughout.

The first part of the question is: what explains (1) and (2)?

My answer is: yes, it is most probably just a curious coincidence, due to:

1. Such groups of two, or more, are common.

2. Modular constrains eliminate most values of $y<10000$.

3. If it were not coincidence, I would expect to see mainly pairs of solutions for higher values of $y$.

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1. Within the $13700$ solutions where $y\not\equiv 0{\pmod 5}$, there is $1$ group of seven, $3$ groups of six, $9$ groups of five, $52$ groups of four, $246$ groups of three, and $1648$ pairs.

In total, $4312$ solutions exist in groups, $>31$ percent, so groups are not rare.

Off topic, I note these give many multiple (4,4,4) solutions.

Off topic, I would personally classify $36$ of these $13700$ solution as (4,1,4) rather than (4,1,5)

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2. Using $n$ for any non-negative integer,

$n^4\equiv 0{\pmod 5}$ when $n\equiv 0{\pmod 5}$

$n^4\equiv 1{\pmod 5}$ when $n\not\equiv 0{\pmod 5}$

$n^4\equiv 0{\pmod {16}}$ when $n\equiv 0{\pmod 2}$

$n^4\equiv 1{\pmod {16}}$ when $n\not\equiv 0{\pmod 2}$

$y^4-x^4\equiv(y-x)(y+x)(y^2+x^2)$

When $y\not\equiv 0{\pmod 5}$ exactly one of the $x_i\not\equiv 0{\pmod 5}$ and the other four $x_i\equiv 0{\pmod 5}$.

Chose $x_1$ to be the one $\not\equiv 0{\pmod 5}$

Dropping the suffix from $x_1$, putting $x=1$, and noting than $y$ must be odd when $x$ is odd (due to $(mod16)$), we have one of, $$(y-1)\equiv 0{\pmod {1025}}$$ $$(y+1)\equiv 0{\pmod {1025}}$$ $$(y^2+1)\equiv 0{\pmod {1025}}$$

In all three cases, we end up with

$(y^4-1)=2^45^4T=10000T$ where $T=w_2^4+w_3^4+w_4^4+w_5^4$ and $(x_i=10w_i)$.

This is necessary, but far from sufficient, for we also need, at least, $T{\pmod {16}}<5$. Indeed, when $T\equiv 0{\pmod {16}}$ this indicates all the $w_i$ are even, and further division(s) of $T$ by $16$ and a corresponding change for $w_i$ are desirable (see $y= 80001$ below).

Back to the three divisors of $y^4-1$.

Just three parametric solutions satisfy $(y^4-1)=10000T$

$y=1250a-1$ gives $(y-1)$ divisible by $1250$

$y=1250a+1$ gives $(y+1)$ divisible by $1250$

$y=625b\pm182$ gives $(y^2+1)$ divisible by $1250$ for odd $b$

The rest of the divisor of $T$, $8$, is always there, but it’s distribution varies between the factors.

With $y<10000$, For the first condition, $(y-1)$, only $a=7,8$ pass the basic modular tests, but produce no solutions.

For $(y+1)$, just $a=5,6,7$ pass the basic modular tests and $a=6$ gives the solutions for $y=7499$

For $(y^2+1)$, just $y=2943,5807,7057,9193$ pass, and $y=9193$ gives solutions.

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3. Solutions for $y\not\equiv 0{\pmod 5}$ with $y<200000$

Red for pairs, blue for groups of three, and green for $gcd(x_i)=20$

3a. Where $(y-1)$ divisible by $1250$

$$\color{red}{72501^4=1^4+62790^4+56940^4+35410^4+6610^4}$$ $$\color{red}{72501^4=1^4+63310^4+56470^4+34320^4+9510^4}$$

$$\color{green}{80001^4=1^4+79480^4+30900^4+19500^4+4520^4}$$

$$\color{red}{107501^4=1^4+102230^4+68500^4+38960^4+9420^4}$$ $$\color{red}{107501^4=1^4+106990^4+39620^4+15460^4+2700^4}$$

$$\color{red}{151251^4=1^4+138060^4+111000^4+47660^4+41890^4}$$ $$\color{red}{151251^4=1^4+141380^4+91600^4+79110^4+61440^4}$$

$$185001^4=1^4+183380^4+65390^4+61020^4+53790^4$$

$$192501^4=1^4+164690^4+150000^4+107030^4+16350^4$$

3b. Where $(y+1)$ divisible by $1250$

$$\color{red}{7499^4=1^4+6350^4+6130^4+3340^4+150^4}$$ $$\color{red}{7499^4=1^4+6970^4+5050^4+3250^4+2520^4}$$

$$29999^4=1^4+26710^4+20410^4+17670^4+13150^4$$

$$\color{blue}{48749^4=1^4+42290^4+35540^4+27280^4+23400^4}$$ $$\color{blue}{48749^4=1^4+46740^4+28600^4+21170^4+8440^4}$$ $$\color{blue}{48749^4=1^4+47950^4+22360^4+18260^4+3240^4}$$

$$\color{red}{54999^4=1^4+46620^4+42560^4+32330^4+15150^4}$$ $$\color{red}{54999^4=1^4+54960^4+11960^4+7850^4+6390^4}$$

$$127499^4=1^4+115370^4+88410^4+68770^4+43660^4$$

$$166249^4=1^4+156830^4+102400^4+81900^4+44750^4$$

$$167499^4=1^4+150890^4+122250^4+76270^4+58320^4$$

$$173749^4=1^4+161050^4+121530^4+65830^4+36140^4$$

3c. Where $(y^2+1)$ divisible by $1250$

$$\color{red}{9193^4=1^4+8680^4+6020^4+3490^4+1410^4}$$ $$\color{red}{9193^4=1^4+8870^4+5410^4+3120^4+920^4}$$

$$15443^4=1^4+14820^4+8720^4+7310^4+460^4$$

$$\color{red}{30807^4=1^4+30120^4+15210^4+12350^4+5500^4}$$ $$\color{red}{30807^4=1^4+30540^4+12660^4+8450^4+2450^4}$$

$$51693^4=1^4+49080^4+31850^4+23380^4+10020^4$$

$$90443^4=1^4+75330^4+65110^4+63680^4+23290^4$$

$$103307^4=1^4+99610^4+55870^4+47580^4+27610^4$$

$$109557^4=1^4+103970^4+69420^4+42270^4+29890^4$$

$$\color{red}{110807^4=1^4+91200^4+86460^4+66270^4+50310^4}$$ $$\color{red}{110807^4=1^4+93930^4+81900^4+72690^4+4560^4}$$

$$114193^4=1^4+112730^4+48750^4+40030^4+24010^4$$

$$136693^4=1^4+122650^4+95550^4+78850^4+30160^4$$

$$142057^4=1^4+120720^4+101710^4+96790^4+16600^4$$

$$\color{red}{148307^4=1^4+139000^4+101560^4+44760^4+16610^4}$$ $$\color{red}{148307^4=1^4+140630^4+95740^4+50680^4+37800^4}$$

$$169193^4=1^4+167150^4+78620^4+28110^4+14000^4$$

$$177943^4=1^4+175730^4+76360^4+55830^4+47840^4$$

$$\color{blue}{190807^4=1^4+166690^4+153280^4+34710^4+1660^4}$$ $$\color{blue}{190807^4=1^4+177140^4+134640^4+55970^4+39530^4}$$ $$\color{blue}{190807^4=1^4+188590^4+83520^4+49670^4+49060^4}$$

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Conclusion. I can’t prove that there is not a fourth power identity for $x=1$, but think that general parametric solutions are more likely.

I’ve tried to avoid excessive detail, but please let me know if any point needs elaboration.

• @Tito Piezas III Perhaps you’ve not noticed my answer? – Old Peter Nov 11 '17 at 16:54