Finding equalities like $2^4+2^4+4^4+3^4+4^4 = 5^4$ using Pythagorean triples?

Question

I. Ramanujan's parameterizations

Ramanujan gave just two quadratic parametrizations to,

$$a^4+b^4+c^4+d^4+e^4 = f^4$$

where $f$ is always integrally divisible by $5$, one of which is,

$$(2x^2+12xy-6y^2)^4+(2x^2-12xy-6y^2)^4+(4x^2-12y^2)^4 \\ +(3^4+4^4)(x^2+3y^2)^4 = 5^4(x^2+3y^2)^4$$

In fact, using a more general identity, we can find infinitely many quadratic parameterizations such as,

$$(50x^2 + 300x y - 150y^2)^4 + (50x^2 - 300x y - 150y^2)^4 + (100x^2 - 300y^2)^4 \\ + (4^4 + 15^4)(x^2 + 3y^2)^4 = 103^4(x^2 + 3y^2)^4$$

as discussed in this post.

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II. Pythagorean triples

Letting $(x,y)=(1,0)$ in Ramanujan's formula above, we get the equality,

$$2^4+2^4+4^4+3^4+4^4=5^4$$

One can notice the Pythagorean triple $(a,b,c)=(3,4,5)$. More generally, it turns out the equation,

$$(p + q)^4 + (-p + q)^4 + (2q)^4 + a^4 + b^4 = c^4$$

where $a^2+b^2=c^2$ is true if the simple condition is satisfied,

$$ab = p^2+3q^2$$

Note: Of course, one can swap $(p,q)\to (q,p)$ depending on the choice of notation.

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III. Question

How do we show there are infinitely many primitive Pythagorean triples $(a,b,c)$ such that,

$$ab = p^2+3q^2$$

and $(p,q)$ are integers?

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P.S. I did a computer search. Examples of the special triple $(a,b,b+1)$ are,

$$2^4+2^4+4^4+3^4+4^4=5^4,\qquad 3^2+4^2=5^2$$ $$2^4 + 32^4 + 34^4 + 13^4 + 84^4 = 85^4,\qquad 13^2 + 84^2 = 85^2$$ $$6^4 + 96^4 + 102^4 + 27^4 + 364^4 = 365^4,\qquad 27^2 + 364^2 = 365^2$$

and so on, though it is unknown if this quadratic-quartic subset has infinitely many members.

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Update: In his answer, Tomita transformed the problem into a family of elliptic curves with an infinite number of rational points. As an bonus, the particular curve he chose with parameter $r=1$ shows an additional structure to the equation, namely,

$$(p + q)^4 + (p - q)^4 + (2p)^4 + a^4 + b^4 = (2p+a)^4$$

where $a^2 + b^2 = c^2 = (2p+a)^2$. Example:

$$4^4 + 2^4 + \color{red}{2}^4 + \color{blue}{3}^4 + 4^4 = (\color{red}{2}+\color{blue}{3})^4 = 5^4$$

$$1506^4 + 2154^4 + \color{red}{648}^4 + \color{blue}{2077}^4 + 1764^4 = (\color{red}{648} + \color{blue}{2077})^4 = 2725^4$$

$$22162738^4 + 22105616^4 + \color{red}{57122}^4 + \color{blue}{128068563}^4 + 3825484^4 \\= (\color{red}{57122} + \color{blue}{128068563})^4 = 128125685^4$$

and so on, for infinitely many primitive Pythagorean triples $(a,b,c)$.

Answer 1

$$ 3p^2+q^2 = 2mn(m^2-n^2) $$
Let $x=\dfrac{m}{n},\ y=\dfrac{q}{n^2},$ and $r=\dfrac{p}{n^2}.$

We have $$y^2 = 2x^3-2x-3r^2.$$ I searched the rational points $(x,y)$ where $p+q<10000$ and $r<100.$

Obviously, this equation can be transformed to Weierstrass form: $$Y^2=X^3-4X-12r^2$$ For the case of $r=1, Y^2=X^3-4X-12r^2$ has rank $1$ and generator is $P(X,Y)=(4,6).$ We get the solutions using group structure $kP$ with $k=1,2,3,4$ as follows.

[r][p+q, p-q, 2p,a,b,c]
[1][4, 2, 2, 3, 4, 5]
[1][1506, 2154, 648, 2077, 1764, 2725]
[1][22162738, 22105616, 57122, 128068563, 3825484, 128125685]
[1][112480990768740, 22759941088740, 89721049680000, 72967048295281, 145407108099600, 162688097975281]
[p+q, p-q, 2p,a,b,c]
[24, 12, 36, 63, 16, 65]
[92, 74, 166, 133, 156, 205]
[106, 58, 164, 133, 156, 205]
[124, 34, 158, 133, 156, 205]
[142, 4, 146, 133, 156, 205]
[210, 150, 60, 675, 52, 677]
[310, 110, 200, 325, 228, 397]
[346, 314, 32, 1443, 76, 1445]
[368, 274, 94, 1443, 76, 1445]
[374, 256, 118, 1443, 76, 1445]
[382, 206, 176, 1443, 76, 1445]
[470, 340, 130, 589, 300, 661]
[534, 438, 972, 637, 1116, 1285]
[666, 282, 948, 637, 1116, 1285]
[610, 50, 560, 925, 372, 997]
[650, 490, 160, 925, 372, 997]
[702, 678, 24, 3843, 124, 3845]
[726, 648, 78, 3843, 124, 3845]
[798, 84, 882, 637, 1116, 1285]
[812, 718, 94, 1333, 444, 1405]
[834, 18, 852, 637, 1116, 1285]
[844, 662, 182, 1333, 444, 1405]
[910, 890, 20, 5475, 148, 5477]
[942, 258, 684, 637, 1116, 1285]
[948, 282, 666, 637, 1116, 1285]
[980, 790, 190, 5475, 148, 5477]
[1046, 862, 184, 1813, 516, 1885]
[1110, 240, 1350, 819, 1900, 2069]
[1114, 626, 488, 1813, 516, 1885]
[1364, 428, 1792, 1273, 2064, 2425]
[1370, 1240, 130, 2451, 700, 2549]
[1386, 798, 2184, 2077, 1764, 2725]
[1740, 570, 1170, 1525, 1548, 2173]
[1772, 364, 1408, 1273, 2064, 2425]
[1770, 780, 990, 1525, 1548, 2173]
[1942, 1838, 104, 4453, 804, 4525]
[1974, 126, 1848, 2077, 1764, 2725]
[2084, 2012, 4096, 3913, 3216, 5065]
[2166, 1464, 702, 2077, 1764, 2725]
[2184, 798, 1386, 2077, 1764, 2725]
[2202, 1266, 936, 2077, 1764, 2725]
[2448, 2286, 162, 5427, 1036, 5525]
[2946, 498, 2448, 3397, 2196, 4045]
[3072, 2154, 918, 3397, 2196, 4045]
[3138, 1842, 1296, 3397, 2196, 4045]
[3356, 356, 3712, 3913, 3216, 5065]
[3382, 3218, 164, 9373, 1164, 9445]
[3894, 3882, 7776, 11557, 3924, 12205]
[3988, 2804, 1184, 3913, 3216, 5065]
[4036, 1412, 2624, 3913, 3216, 5065]
[4044, 114, 4158, 5917, 2844, 6565]
[4326, 492, 3834, 5917, 2844, 6565]
[4604, 4498, 106, 47523, 436, 47525]
[4800, 4680, 120, 50175, 448, 50177]
[4962, 276, 5238, 7957, 3276, 8605]
[5038, 4862, 176, 16093, 1524, 16165]
[5422, 5186, 236, 17653, 1596, 17725]
[5478, 4626, 852, 7957, 3276, 8605]
[5676, 4218, 1458, 7957, 3276, 8605]
[5930, 4820, 10750, 14259, 6100, 15509]
[6576, 6534, 42, 77283, 556, 77285]
[7218, 6114, 1104, 11557, 3924, 12205]
[7446, 5664, 1782, 11557, 3924, 12205]
[7670, 7540, 130, 28525, 2028, 28597]
[7734, 4566, 3168, 11557, 3924, 12205]
[7746, 3282, 4464, 11557, 3924, 12205]
[7878, 4758, 12636, 22477, 5436, 23125]
[8170, 7850, 320, 30589, 2100, 30661]
[8316, 6468, 14784, 9457, 17424, 19825]
[9042, 3498, 5544, 14317, 4356, 14965]

Answer 2

Seiji Tomita has given an Identity on his web site. The link is shown below:

http://www.maroon.dti.ne.jp/fermat/dioph57e.html

Click the fourth power section & select the posting # 86.

The Identity (1), is shown below:

$(-290-500k+40k^2)^4 + (250-80k-290k^2)^4 + (-40-580k-250k^2)^4 + 325^4(1+k+k^2)^4 + 228^4(1+k+k^2)^4 = 397^4(1+k+k^2)^4$

In the above:

$a=228(1+k+k^2)$

$b=325(1+k+k^2)$

$c=397(1+k+k^2)$

$p+q=(250-80k-290k^2)$

$p-q=(-40-580k-250k^2)$

$2q=(-290-500k+40k^2)$

In the above, the condition $ab=p^2+3q^2$ is satisfied so:

$[p-q,\,p+q,\,2q,\,a,\,b]^4=[c]^4$

where $a^2+b^2=c^2$. For $k=0$:

$(290,250,40,325,228)^4=(397)^4$

and $325^2+228^2=397^2$.

Note: Since equation (1) is an identity, it means that that there are many numerical solutions for different values of the parameter 'k'.

Also note that the numerical solutions given by Tito Piezas & Gerry Myerson has the variable 'c' divisible by '5' while the solution in the answer is (not) divisible by '5'. Implying that this identity given in the answer belongs to a different parametric family group.

Note added:

Alternate way to arrive at (4-1-5) quartic equation:

$(q+p)^4+(q-p)^4+(2q)^4+a^4+b^4=c^4$ ------(1)

We take:

$a=(8k-3)$

$b=4(4k-1)(2k-1)$

$c=(32k^2-24k+5)$

the above satisfies, $(c^2=a^2+b^2)$

we have from (1):

$(c^4-a^4-b^4)=(q+p)^4+(q-p)^4+(2q)^4$

After substitution from above we get:

$(c^4-a^4-b^4)=2[(8k-3)(4)(4k-1)(2k-1)]^2 =2(ab)^2$

We also have:

$(q+p)^4+(q-p)^4+(2q)^4=2(p^2+3q^2)^2$

hence: $2(p^2+3q^2)^2=2(ab)^2$

Therefore: $p^2+3q^2=ab$ -- (the condition required)

For, k=2, we get:

$(a,b,c)=(13,84,85)$

from, $(p^2+3q^2)=(ab)=(13*84)=1092$ ----(2)

(p,q)=(3,19) satisfies equation (2)

therefore we get:

$(p,q,a,b,c)=(3,19,13,84,85)$

And substituting above in equation (1) we get:

$(22,16,38,13,84)^4=(85)^4$

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