Solutions to near-perfect Euler bricks?

Question

I. System

We seek to find distinct positive integer solutions $(a,b,c)$ to System 1,

\begin{align} a^2 + b^2 &= \square_1\\[4pt] a^2 + c^2 &= \square_2\\[4pt] b^2 + c^2 &= \square_3 + \color{red}{h^2}\\[4pt] a^2 + b^2 + c^2 &= \square_4 + \color{red}{h^2}\end{align}

Of course, if $h=0$ this is the perfect Euler brick though none are known and, if it exists, then at least one of $(a,b,c) >10^{12}$. As such, we relax our conditions and ask if below a bound on $(a,b,c)$, then what small $h$ can be found? Robert Matson found $h=12$ in a 2015 paper,

$$(a,b,c,h)= (25025, 71820, 5088, 12)$$

However, he was focusing on near-integers,

$$\sqrt{a^2+b^2+c^2} \approx 76225.00094$$

so it is possible he may have ignored smaller $(a,b,c)$. By computer-search, he also found the gigantic,

$$(a,b,c,h)= (117348114345,\, 95932047590764,\, 3644786675612448,\, 24852)$$

where $\sqrt{a^2+b^2+c^2}$ again is a near-integer. Note how small $h$ is compared to $(a,b,c)$, though in his paper, he gives the integer $h^2$ and doesn't seem to have noticed it was a square.

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II. Elliptic curve

Using an elliptic curve, it can be shown that System 1 has infinitely many primitive solutions. However, since Matson's first $(a,b,c)$ is moderately large, the coefficients of the elliptic curve are also large. That is why a smaller solution is desired.

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III. Question

System 1 is equivalent to,

\begin{align} a^2 + b^2 &= \square_a\\[4pt] a^2 + c^2 &= \square_b\\[4pt] a^2 + d^2 &= \square_c\\[4pt] b^2 + c^2 - d^2 &= h^2\end{align}

Q: Can you find four distinct positive integers $(a,b,c,d)$ smaller than Matson's first solution such that the RHS are non-zero squares? With $h$ less than $(a,b,c,d)$ and preferably small. (The solution is necessary for an elliptic curve.)

Answer 1

Method 1

Let $a=2mnk, b=(m^2-n^2)k.$

From systrm of equations $(a^2+c^2 = v^2,\ a^2+d^2 = w^2)$ we get $$(c,d)= \left( \dfrac{mnk(p^2-1)}{p}, \dfrac{4m^2n^2p^2k^2-1}{2p} \right).$$ Since $b^2+c^2-d^2$ must be a square number, then we have quartic equation $$Y^2 = (4m^2n^2k^2-16m^4n^4k^4)p^4+(-8m^2n^2k^2+4k^2m^4+4k^2n^4)p^2+4m^2n^2k^2-1 $$ We searched the rational number $p$ where $(m,n,k)<100$ , height of $p<10000$ and $h<1000.$

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Method 2

Let $a=2mn, b=m^2-n^2.$

From systrm of equations $(a^2+c^2 = v^2,\ a^2+d^2 = w^2)$ we get $$(c,d)= \left( \dfrac{mn(p^2-1)}{p}, \dfrac{mn(q^2-1)}{q} \right).$$ Since $b^2+c^2-d^2$ must be a square number, then we have quartic equation $$Y^2 = m^2n^2q^2p^4+(-m^2n^2q^4+(m^4-2m^2n^2+n^4)q^2-m^2n^2)p^2+m^2n^2q^2 $$ Search range: $(m,n)<100$ , height of $(p,q)<100$ and $h<100.$

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Table

Note: In the table below, three were found by Dmitry Ezhov. Those with an asterisk were found by Method 2.

[h,a,b,c,d]
[3, 79040, 109218, 104013, 150822] (Dmitry)
[5, 3432, 5049, 1615, 5301] (Dmitry)
[5, 29172, 12155, 4896, 13104]*
[7, 360, 378, 1271, 1326]*
[9, 11657240, 5810502, 2050281, 6161622] (Dmitry)
[11, 5520, 6325, 7564, 9860]*
[12, 60, 45, 80, 91]*
[12, 255, 340, 612, 700]
[12, 200, 375, 480, 609]
[12, 1300, 345, 912, 975]
[15, 48, 36, 55, 64]
[15, 156, 117, 65, 133]
[15, 320, 768, 999, 1260]
[15, 140, 147, 171, 225]
[16, 6300, 12480, 10465, 16287]
[20, 936, 1755, 4160, 4515]
[20, 180, 432, 299, 525]
[25, 2160, 228, 871, 900]
[25, 13464, 8670, 5177, 10098]
[26, 360, 150, 325, 357]
[28, 9900, 5075, 9072, 10395]
[30, 136, 102, 255, 273]
[34, 120, 90, 209, 225]
[35, 1584, 1612, 1365, 2112]
[39, 49200, 23985, 20500, 31552]
[40, 1365, 728, 2704, 2800]
[42, 600, 450, 175, 481]
[45, 2772, 429, 560, 704]
[45, 3536, 2652, 2925, 3948]
[52, 420, 832, 851, 1189]
[55, 216, 1287, 1450, 1938]
[55, 8568, 5049, 9350, 10626]
[60, 260, 195, 624, 651]
[60, 696, 3328, 2475, 4147]
[65, 240, 180, 551, 576]
[65, 2772, 1785, 3485, 3915]
[75, 5016, 437, 2880, 2912]
[80, 624, 1980, 1457, 2457]
[81, 4920, 3690, 16031, 16450]
[83, 13860, 1323, 2171, 2541]
[87, 12600, 13230, 31175, 33866]
[90, 1640, 1230, 369, 1281]

(From old table)

[h,a,b,c,d,]
[15, 48, 36, 55, 64]
[65, 240, 180, 551, 576]
[175, 672, 504, 2255, 2304]
[369, 1440, 1080, 6319, 6400]
[671, 2640, 1980, 14279, 14400]
```

Answer 2

Analyzing Tomita's old table above, we find that five solutions,

[a,b,c,d,h]
[48, 36, 55, 64, 15]
[240, 180, 551, 576, 65]
[672, 504, 2255, 2304, 175]
[1440, 1080, 6319, 6400, 369]
[2640, 1980, 14279, 14400, 671]

where $d$ is a square belong to a polynomial parameterization, namely,

\begin{align} a &= 2n(n^2-1)\\[4pt] b &= 3n(n^2-1)/2\\[4pt] c &= (n^2-1)^2-n^2\\[4pt] d &= (n^2-1)^2\\[4pt] h &= n(n^2+1)/2\end{align}

with the five being the odd cases $n=3,5,7,9,11$.

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Update: More generally, if $n\to m/n$, then,

\begin{align} a &= 2mn(m^2-n^2)\\[4pt] b &= 3mn(m^2-n^2)/2\\[4pt] c &= (m^2-n^2)^2-m^2n^2\\[4pt] d &= (m^2-n^2)^2\\[4pt] h &= mn(m^2+n^2)/2\end{align}

For example, if we use $n=2$ and $m=5,7$, we get,

[h,a,b,c,d]
[145, 420, 315, 341, 441]
[371, 1260, 945, 1829, 2025]

found in a table with $h<1000$, and so on.