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An Euler brick is a triple of positive integers $(a, b, c)$ with $a\leq b\leq c$ such that $a^2+b^2, a^2+c^2, b^2+c^2$ are all squares. We call $a, b, c$ the side lengths of said Euler brick.

As a follow-up to this question, I am wondering about the existence of distinct Euler bricks with the same smallest side length. In other words, do there exist Euler bricks $(a, b, c)$ and $(a, b', c')$ where $b\neq b'$ and $c\neq c'$? How many distinct Euler bricks can exist with a fixed smallest side?

Bill Dubuque
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Mathieu Rundström
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    Take any 2 Euler bricks $(a, b, c)$ and $(A, B, C)$, then $(aA, bA, cA)$ and $(aA, aB, aC)$ are Euler bricks with the same smallest length. $\quad$ We can extend this approach to $n$ Euler bricks, and create $n$ Euler bricks with the same smallest length. – Calvin Lin Nov 22 '24 at 01:44
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    If you insist on primitive Euler bricks (that is, $\gcd(a,b,c)=1$, there are (at least) two with smallest side $1008$. See https://oeis.org/A031175 – then from https://oeis.org/A031173/a031173.txt we can see they are $(1155, 1100, 1008)$ and $(12075, 1100, 1008)$. – Gerry Myerson Nov 22 '24 at 03:15
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    There are also two at $2964$. – Gerry Myerson Nov 22 '24 at 03:21
  • @CalvinLin It seems there are infinitely many pairs of semi-primitive Euler bricks (one brick may have all even dimensions) that share two sides. Kindly see answer below. – Tito Piezas III Jan 01 '25 at 16:26

1 Answers1

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There are Euler bricks that share one side (the smallest). As pointed out by Myerson, an example is,

$$(a,b,c) = (\color{blue}{2964}, 9152, 9405)\\ \,\,(a,b,c) = (\color{blue}{2964}, 6160, 38475)$$

However, if we relax your question and allow two sides, then there are examples as well,

$$(a,b,c) = (1008, 1100, 1155)\\ \,\,(a,b,c) = (1008, 1100, 12075)\\ (a,b,c) = (1008, 1100, \color{red}{960})\; $$

though the third has a smaller side. For the case of two sides, there are infinitely many given by a quadratic form.

If $u^2+v^2=5w^2,$ then,

$$(a,b,c) = \big[(u^2-w^2)(v^2-w^2)/2,\; 2uvw^2,\; uw(v^2-w^2)\big]\\ (a,b,c) = \big[(u^2-w^2)(v^2-w^2)/2,\; 2uvw^2,\; vw(u^2-w^2)\big]$$

Note that we can swap $(u,v)$ and $(a,b)$ remains unchanged, as it is only $c$ that changes. For some $(u,v,w)$, it may yield a shared side that is not the smallest. Recall the above,

Let $(u,v,w) = (2,11,5)$, then

$$(a,b,c) = (1008, 1100, \color{red}{960})\\ \;\;(a,b,c) = (1008, 1100, 1155)$$

However, there are appropriate $(u,v,w)$.

Let $(u,v,w) = (19,22,13)$, then

$$(a,b,c) = (30240, 141284, 77805)\\ (a,b,c) = (30240, 141284, 54912)$$

Let $(u,v,w) = (22,31,17)$, then

$$(a,b,c) = (65520, 394196, 251328)\\ (a,b,c) = (65520, 394196, 102765)$$

and infinitely more. A subset involves transforming our quadratic form,

$$u^2+v^2=5w^2$$

Let $u= \dfrac{t+3}2$ and $v=\dfrac{3t-1}2$ to get the well-known Pell equation,

$$t^2-2w^2=-1$$

where $t = 1,7,41,239,1393,\dots$ (A002315) and $w = 1, 5, 29, 169, 985,\dots$ (A001653) though we start with $t>7$ to get $(a,b,c)\neq0$. For example, let $t=41$ then,

$$(a,b,c) = (514080, 2257244, 1837440)\\ (a,b,c) = (514080, 2257244, 631533)\;\,$$

and so on for infinitely many $t$.

P.S. Note that $x = u-2= \dfrac{t+3}2-2 = 0, 3, 20, 119, 696,\dots$ (A001652) such that $x^2+(x+1)^2 = w^2$, while $v=\dfrac{3t-1}2 = 1, 10, 61, 358, 2089,\dots$ is (A301383).

Tito Piezas III
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    Nice! Now it only remains to show that there are infinitely many sharing three sides! ;) – Mathieu Rundström Jan 01 '25 at 17:41
  • @MathieuRundström A similar "pairing" phenomenon occurs for Euler quadruples $(a,b,c,d)$ where $a^2+b^2+c^2 = x_1^2$, etc, and one can find pairs of quadruples that share two variables. – Tito Piezas III Jan 02 '25 at 03:58