Euler bricks (or Euler cuboids)
Christian Boyer, France, April-May 2012
(May 2020, short update with Matson's computing results published in 2015)

Even if I think that 3D perfect Euler bricks and 4D Euler bricks do not exist... here are some attempts...
Interesting results are bordered.
Open questions are in green. Feel free to send me remarks.

 In this page: An example of a 4D "diabolic" near-solution (693, 140, 480, 2376),giving 7 squares out of 8, as detailed and explained in this page: 693² + 140² = 707²       is a square 693² + 480² = 843²       is a square 140² + 480² = 500²       is a square 693² + 2376² = 2475²    is a square 140² + 2376² = 116² * 421  is not a square... unfortunately... 480² + 2376² = 2424²    is a square 693² + 140² + 480² + 2376² = 2525²    its 4D diagonal is a square! 693 * 140 * 480 * 2376 = 332640²       the product of its edges is also a square!

"Near-solution" means only one missing equation. "Near-miss" means that the integer is close to a squared integer in this only missing equation.

3D Euler bricks

• a² + b² = p²          [system 1]
• a² + c² = q²
• b² + c² = r²

Numerous solutions are known. For example, from his family of solutions, Euler provided (117, 44, 240):

• 117² + 44² = 125²
• 117² + 240² = 267²
• 44² + 240² = 244²

See Wikipedia or Weisstein-MathWorld-Wolfram.

• any primitive 3D Euler brick (a, b, c) has one odd edge (say a) and two even edges (implies b and c).
• any primitive 3D Euler brick (a, b, c) has a derived 3D Euler brick (ab, ac, bc).

3D perfect Euler bricks

• a² + b² = p²          [system 2]
• a² + c² = q²
• b² + c² = r²
• a² + b² + c² = s²

This system adds the 4th equation in bold to the 3D Euler brick problem: the internal diagonal has to be an integer.
It is not proved that this problem is impossible, but no solution is known: see Weisstein-MathWorld-Wolfram, this unsolved problem by Tim Roberts, and the interesting searches by Randall Rathbun, by Bill Butler and by Rob Matson.
For example the 3D Euler solution (117, 44, 240) is not perfect because 117² + 44² + 240² = 73225 is not a squared integer: 73225 = 5² * 29 * 101.

If Rathbun's computing is correct, there is no solution with min(a, b, c) < 2.325e+10.
If Butler's computing is correct, there is no solution with odd edge a < 3e+12.
If Matson's computing is correct, there is no solution with odd edge a < 2.5e+13.

 If my computing is correct: no solution with s < 1e+15, s having all its prime factors < 1000. no solution with s < 1e+15, s having 5 (or less) prime factors f1<1000000, f2<10000, f3<1000, f4<100, f5<100. no solution with s < 1e+17, s having its bigger prime factor f1 < 1000 and its other prime factors < 100. My searching method is briefly described here.

We can notice that the system 2 is a particular case of this other system, when k = 0:

• a² + b² = p² + k          [system 3]
• a² + c² = q² + k
• b² + c² = r² + k
• a² + b² + c² = s² + k

It is possible to obtain solutions with some small k? Yes! Here is my nearest miss, a VERY nice solution (a, b, c, k) = (1887, 2943, 5247, 2). With a so small k, this is very close to a 3D perfect Euler brick!

 1887² + 2943² = 3496² + 2           = (3496.00028...)²             1887² + 5247² = 5576² + 2           = (5576.00017...)²             2943² + 5247² = 6016² + 2           = (6016.00016...)² 1887² + 2943² + 5247² = 6305² + 2           = (6305.00015...)²

I have found only three solutions with -10000 < k < 10000, the example above and these two others:

• (7436, 52197, 62757, 729)
• (333, 364, 525, -6615)

Solutions with small k are very rare. Is it possible to obtain other primitive solutions of [system 3] with abs(k) < 10000?

How close to an integer can the internal diagonal be? Any 3D Euler brick [system 2] can be written that way, s being the integer minimizing abs(k), with k positive or negative:

• a² + b² = p²          [system 4]
• a² + c² = q²
• b² + c² = r²
• a² + b² + c² = s² + k

Because we would like k = 0, is it at least possible to obtain some small k? I think that the nearest miss, the best known brick having the smallest abs(k), is this one, with k = -120:

• (85, 132, 720) -> 85² + 132² + 720² = 737² - 120

I have found only six primitive 3D Euler bricks with -1000 < k < 1000, the example above and these five other near misses:

• (117, 44, 240) -> 117² + 44² + 240² = 271² - 216 = 270² + 325
• (195, 748, 6336) -> 195² + 748² + 6336² = 6383² - 264
• (855, 832, 2640) -> 855² + 832² + 2640² = 2897² + 240
• (28083, 43056, 105820) -> 28083² + 43056² + 105820² = 117645² + 400
• (37835, 269280,1244484) -> 37835² + 269280² + 1244484² = 1273846² - 165

Solutions with small k are very rare. Is it possible to obtain other primitive solutions of [system 4] with abs(k) < 1000?

The bigger the 3D bricks, the bigger their k. For example, with any s > 1e+8, it seems extremely difficult (or impossible?) to obtain abs(k) < 1e+5, the best solution being probably:

• (92951815, 58369920, 87095808) -> 92951815² + 58369920² + 87095808² = 140116977² + 154960

Is it possible to obtain primitive solutions of [system 4] with s > 1e+8 and abs(k) < 1e+5?

And we want k = 0... The bigger the 3D bricks, the smaller will be the probability to directly obtain k = 0, the perfect Euler brick seems hopeless...

4D Euler bricks

• a² + b² = p²          [system 5]
• a² + c² = q²
• b² + c² = r²
• a² + d² = s²
• b² + d² = t²
• c² + d² = u²

Today, it is unknown if a 4D Euler brick (a, b, c, d) can exist: see this unsolved problem by Tim Roberts. If they exist:

• any primitive 4D Euler brick (a, b, c, d) has one odd edge (say a) and three even edges (implies b, c and d).
• any primitive 4D Euler brick (a, b, c, d) has a derived 4D Euler brick (abc, abd, acd, bcd).

General idea on how to obtain 4D near-solutions

A 4D brick can be seen as a combination of three 3D Euler bricks having the same odd edge a:

• a 3D Euler brick (a, b, c)
• a 3D Euler brick (a, b, d)
• a 3D Euler brick (a, c, d)

An idea which could solve this problem is to combine two primitive 3D Euler bricks having the same ratio of two edges, then scale them in order to obtain the same edges: this will directly produce a solution of at least 5 out of the 6 equations of the system 5! If the 6th equation with the two remaining edges is not true, we will call it a "4D near-solution". If the equation is true, then we have the Grail, a 4D solution.

For example the 3D Euler solution (117, 44, 240) and this other 3D solution (429, 880, 2340) have one same ratio 117 / 240 = 429 / 880.
We will scale theses bricks by factors 11 and 3, because 11*117 = 3*429 = lcm(117, 429) =
1287.
Now 11*(117, 44, 240)=(
1287, 484, 2640) and 3*(429, 880, 2340)=(1287, 2640, 7020) have two edges in common in red and in blue, and we obtain the 4D near-solution (1287, 484, 2640, 7020).
Checking the only missing equation of the two other edges in italics, unfortunately this do not give a squared integer:

• 1287² + 484² = 1375²
• 1287² + 2640² = 2937²
• 484² + 2640² = 2684²
• 1287² + 7020² = 7137²
• 484² + 7020² = 49514656 = 2^5 * 293 * 5281
• 2640² + 7020² = 7500²

Three 4D near-solutions from any 3D brick, using the derived brick

The properties of the previous example (1287, 484, 2640, 7020) can be explained because we have combined a brick and its derived brick. The derived brick of (117, 44, 240) is (429, 880, 2340), because (117*44, 117*240, 44*240) = (5148, 28080, 10560) = 12*(429, 2340, 880). With this brick and its derived brick, we can in fact generate three 4D near-solutions! The two others are (429, 880, 2340, 4800), and (4563, 1716, 3520, 9360). Why three?

 Each primitive 3D brick (a, b, c), combined with its derived brick (ab, ac, bc), produces three 4D near-solutions having at least 5 true equations out of 6: a*(a, b, c)=(a², ab, ac)  &  (ab, ac, bc)  ->  (ab, ac, bc, a²) which could become a 4D Euler brick if and only if a4 + b²c² is a squared integer (the 6th missing equation) b*(a, b, c)=(ab, b², bc)  &  (ab, ac, bc)  ->  (ab, ac, bc, b²) which could become a 4D Euler brick if and only if b4 + a²c² is a squared integer (the 6th missing equation) c*(a, b, c)=(ac, bc, c²)  &  (ab, ac, bc)  ->  (ab, ac, bc, c²) which could become a 4D Euler brick if and only if c4 + a²b² is a squared integer (the 6th missing equation) Maybe one of the known parametric 3D bricks (a, b, c) will allow a4+b²c² or b4+a²c² or c4+a²b² to be a square, and then immediately produce a 4D brick?For example the parametric solution (x(x²-3y²), y(3x²-y²), 4xyz), with x=p²-q², y=2pq, z=p²+q² produces Euler's solution (117, 44, 240) when p=2, q=1. Can [x(x²-3y²)]4 + [4xy²z(3x²-y²)]² be a square?

Proof. If (a, b, c) is a 3D Euler brick, then (ab, ac, bc, a²) is a 4D near-solution because:

• (ab)² + (ac²) = a²(b²+c²) = (ar)² is a square
• (ab)² + (bc²) = b²(a²+c²) = (bq)² is a square
• (ab)² + a4 = a²(b²+a²) = (ap)² is a square
• (ac)² + (bc²) = c²(a²+b²) = (cp)² is a square
• (ac)² + a4 = a²(c²+a²) = (aq)² is a square
• (bc)² + a4 is unknown

Similar proofs with (ab, ac, bc, b²) and (ab, ac, bc, c²) which are the two other near-Euler bricks.

Three more 4D near-solutions from any 3D brick, repeating an edge

 Each primitive 3D brick (a, b, c) produces three 4D near-solutions, in simply repeating one of its edges: (a, b, c, a) (a, b, c, b) (a, b, c, c) Unfortunately none of these supplemental solutions can be a 4D brick, because twice a square cannot be a square.

However here a nice solution (a, b, c, c), very close to a 4D Euler brick. This is the nearest possible miss, a delta of only 1:

(4901, 4368, 13860, 13860)
13860² + 13860² = 2*(13860)² = 19601² - 1 (= 19600.999974...)²

Other 4D near-solutions

Using derived bricks or repeating edges are not the only ways to obtain near-solutions. We can also obtain 4D near-solutions combining bricks and NOT-derived bricks, for example :

17*(117, 44, 240) & 4*(187, 1020, 1584) -> (1989, 748, 4080, 6336)
1989² + 748² + 4080² + 6336² = 61306921 = 13 * 761 * 6197

Another 4D near-solution, with something strange, two odd edges instead of only one...:

3*(117, 44, 240) & (85, 132, 720)  ->  (85, 351, 132, 720)
85² + 351² = 130426 = 2 * 65213

Computing

 Using the 30414 primitive 3D solutions with smallest edge < 2.325e+10 (*) kindly provided by Randall Rathbun, and their derived solutions (even if their smallest edges become > 2.325e+10), there was a (small) hope to obtain a 4D solution. Combining them, excluding solutions repeating edges which cannot produce the 6th equation, I have constructed 93550 couples having a same ratio. Here is a zipped text file (6Mb) of 93550 primitive 4D near-solutions. By construction, all of of these solutions have 5 equations true, but unfortunately, none of them produce the 6th equation true...

Among these solutions, here is the nearest miss of a squared integer, with a delta of only 96. This solution do not come from a brick and its derived brick:

21*(101565, 240900, 1041392) & 61*(34965, 62900, 358512)  ->  (2132865, 3836900, 5058900, 21869232)
3836900² + 5058900² = 40314270820000 = 2^5 * 5^4 * 16657 * 121013 = 6349352² + 96

If somebody has a longer list of primitive 3D solutions, maybe this method will produce a 4D solution?

(*) Meaning that other edges are bigger, the biggest edge of this list being ~2.2e+15

4D perfect Euler bricks

• a² + b² = p²          [system 6]
• a² + c² = q²
• b² + c² = r²
• a² + d² = s²
• b² + d² = t²
• c² + d² = u²
• a² + b² + c² + d² = v²

This system adds the 7th equation in bold to the 4D Euler brick problem: the 4D internal diagonal has to be an integer. Because we don't know a solution of the 6 first equations, obviously we do not know a solution to these 7 equations. But if we analyze the first numerical example (1287, 484, 2640, 7020) of the previous part, the 7th equation is true!

• 1287² + 484² + 2640² + 7020² = 7625²

Surprising, but in fact easy to explain. Because this solution and a lot of others come from a combination of bricks and their derived bricks, we have seen that their form is (ab, ac, bc, a²) or similar. The sum of squares of their edges is a²b² + a²c² + b²c² + a²a²  = (a²a² + a²b²) + (a²c² + b²c²) = a²(a² + b²) + c²(a² + b²) = (a² + c²)(a² + b²) = (pq)², i.e. a square!

Among the 93550 4D near-solutions computed in the previous part, there are 71553 primitive 4D perfect near-solutions (5Mb zipped text file). Here are the smallest perfect near-solutions, with 4D diagonals v < 10000:

• v = 2117 with (1155, 960, 1008, 1100)
• v = 2525 with (693, 140, 480, 2376)
• v = 5429 with (429, 880, 2340, 4800)
• v = 7025 with (1617, 1120, 3840, 5544)
• v = 7625 with (1287, 484, 2640, 7020)
• v = 8845 with (187, 1020, 1584, 8640)
• v = 9577 with (935, 1452, 7920, 5100)

The nearest miss of a squared integer that I have found is with (693, 140, 480, 2376), delta of 576:

• 140² + 2376² = 5664976 = 2^4 * 29^2 * 421 = 2380² + 576

4D "diabolic" Euler bricks

• a² + b² = p²          [system 7]
• a² + c² = q²
• b² + c² = r²
• a² + d² = s²
• b² + d² = t²
• c² + d² = u²
• a² + b² + c² + d² = v²
• abcd = w²

This system adds a 8th equation to the 4D perfect Euler brick problem: the product of the 4 edges has to be a square. Excellent news: solutions with 7 true equations out of 8 are very easy to find. All of our 7 examples with v < 10000 listed above have this last equation true! For example with (1287, 484, 2640, 7020):

• 1287 * 484 * 2640 * 7020 = 3397680² = (2^4 * 3^3 * 5 * 11^2 * 13)²

Looks surprising, but again easy to explain. Because these solutions come from combinations of bricks and their derived bricks, their form is (ab, ac, bc, a²) or similar. The product of their edges is ab * ac * bc * a² = (a²bc)², i.e. a square!

Here are listed the full nice characteristics of (693, 140, 480, 2376), one of these 4D "diabolic" near-solutions:

 693² + 140² = 707² 693² + 480² = 843² 140² + 480² = 500² 693² + 2376² = 2475² 140² + 2376² = 116² * 421 480² + 2376² = 2424² 693² + 140² + 480² + 2376² = 2525² 693 * 140 * 480 * 2376 = 332640²

In fact, it's very difficult to find 4D perfect near-solutions having a² + b² + c² + d² = v² where abcd is not a square: there are only two examples among the list of 71553!

• 24515225 * 31124280 * 83023920 * 55040832 = 3486770975373444337997214720000 = 2^13 * 3^6 * 5^4 * 7^3 * 11^2 * 17^4 * 19^3 * 73^3 * 101
• 55257111 * 1361579520 * 851466000 * 2275697600 = 145785069323664796320499556352000000 = 2^19 * 3^4 * 5^6 * 7^3 * 11^2 * 13 * 19^3 * 31^2 * 43^2 * 53 * 67 * 97^2

•  In the list of 71553 primitive 4D perfect near-solutions, given in the previous part, 71551 of them are "diabolic" near-solution, with 7 true equations out of 8.

It's also very difficult to find 4D near-solutions having abcd = w² where a² + b² + c² + d² is not a square. Because I am unable to find bigger examples, here are perhaps the two only possible examples. They are derived, because with (a, b, c, d) = (1105, 9360, 29700, 35904), we can construct (abc, abc, acd, bcd) = (1105*9360*29700, 1105*9360*35904, 1105*29700*35904, 9360*29700*35904) = 105019200*(95040, 11220, 3536, 2925).

• 1105² + 9360² + 29700²  + 35904² = 2260017841 = 97 * 193 * 120721
• 2925² + 3536² + 11220² + 95040² = 9179548921 = 617 * 14877713

Just for fun, here the biggest 4D "diabolic" near-solution of my long list: (15607515986148064713435, 68420270979732808900, 821525407157461544149920, 187400177485923053458405368).

4D fully perfect Euler bricks

The 4D "perfect" Euler bricks check only its 2D and 4D diagonals. If we also want the 3D diagonals of 4D bricks, we will call them "fully perfect" bricks and we will have to solve:

• a² + b² = p²          [system 8]
• a² + c² = q²
• b² + c² = r²
• a² + d² = s²
• b² + d² = t²
• c² + d² = u²
• a² + b² + c² = v²
• a² + b² + d² = w²
• a² + c² + d² = x²
• b² + c² + d² = y²
• a² + b² + c² + d² = z²

Hmmm.... Because we do not know any 3D perfect brick, we do not know any 4D near-solution having at least p², q², r² and v².

5D (or 6D or 7D...) Euler bricks

Again hmmm... However, with similar ideas combining 3D Euler bricks, we may find solutions with some of the needed equations which will be true.

For example, with any 3D Euler brick (a, b, c), we can construct a 6D (ab, ac, bc, a², b², c²) which will have at least 9 equations true. With Euler's (117, 44, 240), we obtain (5148, 28080, 10560, 13689, 1936, 57600):

•   5148² + 28080² = 28548²
•   5148² + 10560² = 11748²
•   5148² + 13689² = 14625²
•   5148² +   1936² =  5500²
• 28080² + 10560² = 30000²
• 28080² + 13689² = 31239²
• 28080² + 57600² = 64080²
• 10560² +   1936² = 10736²
• 10560² + 57600² = 58560²

But unfortunately the 6 other sums are not squares:

•   5148² + 57600²
• 28080² +   1936²
• 10560² + 13689²
• 13689² +   1936²
• 13689² + 57600²
•  1936² + 57600²