...

a short theorem: if 2E >= 3V, then Hadwiger's conjecture is true:: proof: assume 2E < 3V; if

we let V= 4, then E < 6, and the complete graph, K4, could have no more than 5 edges and

wouldn't require 4 differently-colored vertices. if 2E >= 3V, then using Euler's characteristic,

2V -2E +2F = 4, and by substitution 2F -V >= 4; also, 6F -3V >= 12, -2E < -3V, & 3F -E >= 6;

thus, {3F -E >= 6} minus {2F -V >= 4} implies that V -E +F >= 2; thus, if the graph is "at least"

planar, V >=4, and 2E >=3V, then Hadwiger's conjecture is absolutely valid. of course, we

are assuming that the graph is simply connected; the one {4-vertex, 6-edge} planar graph is

K4, the two non-planar graphs are both a square with 2 diagonals that cross but do not form

a vertex and a solid tetrahedron! or... we'd have some multiple of these three cases!

*QED 03/4/2012

...

...

I invited a mathematcian to view the next two proofs, and he said that they were not rigorous?

I told him that they contained equations that were only manipulated using algebra that an 8th-

grader could understand, and then I asked him to stop contacting me; obviously, he couldn't

read... he still contacted me and could not elaborate on why he didn't not like my derivations.

I attended a lecture by Dr. Paul Halmos (who died in 2006); he insisted that we fight the math

problems... not the eventual answers!

...

the minimum crossing number, cr(K[m,n]) of a complete bipartite graph by me for Zarankiewicz

as Cauchy did for Euler's topological characteristic formula, V -E +F = 2.

...

cr(K[m,n])= floor(m/2)*floor((m-1)/2)*floor(n/2)*floor((n-1)/2)

...

proof:

draw n (even) points in two parallel rows; draw all possible lines from each point to the rest.

divide by 2 to remove the duplicate connections with regard to orientation and divide by 2

again to count only those lines from a point in one row to the others excluding the point di-

rectly opposite the ones in question. thus, we have K(hat)[n,n]= floor(n/2)*floor((n-1)/2) con-

nections; just imagine that we don't connect the points that are immediately adjacent.

...

now, let 'm' equal the number of points in the top row and 'n' equal the number of points in the

bottom row. since two lines and only two lines are allowed to produce a crossing, we can sub-

stitute floor(m*n)/2 in for each 'n' in the 'connection-counting' transformation.

this results in two functions that describe the number of crossings for a set of points in two par-

allel rows. thus, cr(K[m,n])= min{floor(m/2)*floor(n/2)*floor((m*n-1)/4)}= min{floor(m/2)*floor(n/2)

*floor((m+1)/2)*floor((n-1)/2), floor(m/2)*floor(n/2)*floor((m-1)/2)*floor((n+1)/2)}= floor(m/2)*floor

((m-1)/2)*floor(n/2)*floor((n-1)/2) since (m*n-1) is actually the difference of two squares where 'm'

and 'n' are half of the original 'n' points.

...

thus, we have cr(K[m,n])= floor(m/2)*floor((m-1)/2)*floor(n/2)*floor((n-1)/2) as the supremum or

unique minimum of the topological object in question.

...

therefore, cr(K[m,n])= floor(m/2)*floor((m-1)/2)*floor(n/2)*floor((n-1)/2).

*QED

...

...

the minimum crossing number, cr(K[n,n]) of a complete graph by me for Richard K. Guy who's 94

years old at the rendering date of this proof to him. initially, he and his protege said that they

weren't interested in the proof, but I noticed that someone from Calgary visited my website on the

very next day... yeah, right...

...

cr(K[n,n])= (1/4)*floor(n/2)*floor((n-1)/2)*floor((n-2)/2)*floor((n-3)/2)

...

proof:

use K(hat)[n,n]= floor(n/2)*floor((n-1)/2) to count the number connections regardless of orientation

and considering that points must be placed in 2 concentric circles to achieve the desired result. re-

place 'n' with 'floor(n*(n-2))/4' to count the total number of crossings in a complete graph, spanning

the set from selected points; thus, cr(K[n,n])=min{(1/4)*floor(n/2)*floor((n-2)/2)*floor((n*(n-2)-1)/4)}=

min{floor(n/4)*floor((n-2)/4)*floor(((n-1)*(n-1))/4), floor(n/4)*floor((n-2)/4)*floor((n-3)*(n-1))/4)}=(1/4)*

floor[(n/2)*((n-1)/2)*floor((n-2)/2)*floor((n-3)/2)] where the polynomial 'n*(n-2)-1' can be manipulated

since it's closely divided by 4 within its own floor function. hence, the supremum becomes...

...

cr(K[n,n])= (1/4)*floor(n/2)*floor((n-1)/2)*floor((n-2)/2)*floor((n-3)/2).

*QED

...

NOTE: fewer partitions would mean that we couldn't count using the crossing-transform because it

wouldn't exist, and additional partitions would increase the number of multiplied, floor functions!

the arguments that I'm receiving about the minimum crossings' proofs is that I'm only coming from

above and not from below. that isn't true. I'm coming from above by using the transform floor(n/2)*

floor((n-1)/2) to produce "average counting functions" and selecting a supremum which is unique

by its very definition. the only way to come from below is to divide by '3' or some larger number

in either the transform or the spanning portions, and that would give you the wrong counts!

...

it's very, very difficult to believe that either Zarankiewicz or Guy could have come up with either of

the supremums of two of the minimum counting functions without prior knowledge of the functions

themselves; it's definitely not a simple question or answer, but both persons just happened to pick

the 'needle-out-of-a-haystack' for answers from a long list of careful observations from an even long-

er list of articulated drawings; that's too much work!

...

...

Proth's theorem extended (not by much, but w/out sacrifice!):

...

let Q= k*2^n +1, n >2 is a natural number and k <= 2^n+1. If for some 'a', a^((Q-1)/4) == +/-1 (mod Q),

then 'Q' is prime. (this theorem lays hidden unless also gcd(n, 3) = 1.)

...

proof:

if 'm' is from the set of natural numbers, then every odd prime divisor 'q' of a^(2^(m+1))+/-1 implies

that q == +/-1(mod a^(m+2)) [concluded from generalized Fermat-number 'proofs' by Proth and ad-

justed by me replacing 'm' with 'm+1'].

...

now, if 'p' is any prime divisor of 'R', then a^((Q-1)/4) = (a^k)^(2^(n-2)) == +/-1(mod p) implies that p

== +/-1 (mod 2^n).

...

thus, if 'R' is composite, 'R' will be the product of at least two primes each of which may have a maxi-

mum value of (2^n +1), and it follows that...

...

k*2^n +1 >= (2^n +1)*(2^n +1) = (2^n)*(2^n) + 2*(2^n) +1; but the 1's cancel, so k*(2^n) >= (2^n)*(2^n)

+2*(2^n) and upon dividing by 2^n... implies that k>= 2^n +2.

...

hence, for some 'a', if k <= 2^n +1 and a^((Q-1)/4) == +/-1 (mod Q), then 'Q' is prime.

*QED

...

this simple improvement isn't significant enough to be published, but since I was able to squeeze a

little more out of his 1878-proof, it became the idea that persuaded me to look at other proofs.

...

...

(my) Riesel's theorem (as it should be named):

...

let R= k*2^n -1, n >2 is a natural number and k <= 2^n -1. If for some 'b', b^((R-1)/2) == +/-1 (mod R),

then 'R' is prime; (similar to Proth's theorem, and this theorem lays hidden unless also gcd(n, 3) = 1.)

...

proof:

if 'm' is from the set of natural numbers, then every odd prime divisor 'r' of a^(2^m) +1 implies that q

== +/-1(mod a^(m+1)) [concluded from generalized Fermat-number 'proofs' by Proth, and after my ex-

amination of Proth's theorem].

...

now, if 'q' is any prime divisor of 'S', then b^((R-1)/2)= (b^k)^(2^(n-1)) == +1 (mod q) implies that q

== +/-1 (mod 2^n).

...

thus, if 'S' is composite, 'S' will be the product of at least two primes each of which may have a maxi-

mum value of (2^n -1), and it follows that...

...

k*2^n +1 = (2^n)*(2^n) -1 = (2^n +1)*(2^n -1) <= (2^n)*(2^n) -2*(2^n) +1; implies that -2 <= -2*(2^n)

and dividing by 2^n and multiplying by -1, we have.. 1 > 2^n, contradiction!

...

hence, for some 'b', if k <= 2^n -1 and b^((R-1)/2) == +/-1 (mod R), then 'R' is prime.

*QED

...

Note: a Mersenne number can be easily checked for primality using only one Riesel theorem test!

...

...

some Mersenne divisors theorems:

(all of them can be proved similar to this one)

...

I. theorem: if k >1 and p= 4k +1 is 'prime', then 2*p +1 is also 'prime' iff 2*p +1 divides Mp +2;

a 'complimentary' proof to that of Euler's result for mersenne divisors [shown by modifying La-

grange's proof of the original if p= 4k +3,...].

...

proof(forward): suppose that q= 2*p +1 is prime; q= 3 (mod 8) and 2 is a quadratic residue

modulo q, and there exists an integer 'n' such that n^2= 4 (mod q). Thus, 2^p = 2^((q-1)/2) =

n^(q-1) = -1 (mod q); adding 2 to each portion of this equation shows that (2*p +1) | (Mp +2).

...

proof(backward): let 2*p +1 be a composite factor of Mp +2 & let 'q' be the least prime factor;

2^p= 1 (mod q) and the order of 2 divides both p and q-1; so, the conclusion is the same as

the original Lagrange proof, (2*p +1) +1 > q^2 > p^2 contradicts that p> 2.

*QED

...

II. if p= 4k +3 is 'prime', then 2*p +1 is also 'prime' iff 2*p +1 divides Mp, or 2^p== 1 mod (2p +1).

...

III. if k >1 and p= 4k +1 is 'prime', then 2*p +3 is also 'prime' iff 2*p +3 divides Mp -p, or 2^p -p==1

mod (2p +3).

...

IV. if p= 4k +3 is 'prime', then 2*p +3 is also 'prime' iff 2*p +3 divides Mp -p -1, or 2^p -p == 2

mod (2p +3).

...

...

this proof needs to be corrected. I'm in the process of doing so...

...

Olsen's conjecture is actually a theorem. his conjecture is equivalent to the following as posted

on the Open Problem Garden by M. DeVos. for every finite multiplicative group G, let z(G) denote

the smallest integer 'm' such that every sequence of 'm' elements of G has a sub-sequence of length

|G| with a product equal to 1 in the given order (so Olson's conjecture is equiv. to z(G) <= 2*|G| -1).

it is clear that z(G) <= |G|(|G| -1) +1, since any sequence of length > |G|(|G| -1) must contain at least

|G| copies of the same element, and the product of these will be 1.

...

this statement is equivalent to... if... z(G) <= |G|(|G| -1) +1, then... z(G) <= 2*|G| -1.

it's very true... no one just put pencil to paper; it can be found on the Open Problem Garden!

...

proof: assume that #1.) z(G) > 2|G| -1; we know comfortably that #2.) z(G) <= |G|(|G| -1) +1, and

solving for #1.), we have... [z(G) +1]/2 > |G|. so, by subst. of #1.) into #2.), #2.) becomes z(G) <=

([z(G) +1]/2)*(([z(G) +1]/2) -1) +1. so, 4*z(G) <= (z(G))^2 +2*z(G) +1 -2*z(G) -2 +1 <= (z(G))^2,

or... 4*z(G) <= (z(G))^2... and... z(G) >= 4. and since #1.)... 5/2 >= |G| implies that |G| = 1 or |G| =

2, but by #2.), either z(G) <= 1 or z(G) <= 3; both equalities from #1). are contradictory to #2.),

z(G) >= 4. hence, by contradiction... z(G) <= 2|G| -1.

*QED

01/29/2013

...

...

finally,...

a perfect cuboid cannot exist; I don't know why this problem has presumably existed for so

long; here are the closest solutions...

...

honorable mention:

...

if two non-planar rectangles share an adjacent integral edge and both have integral diagonals,

then if that shared edge were to be a part of an integral cuboid, then those rectangular faces

cannot produce an integral main-cube diagonal or an integral diagonal for the remaining lateral-

or end-face, however you see it...

...

the stereo-typical result would be a cuboid with edges 3, 4, and 4.

...

runner-up:

...

if two non-planar triangles share an adjacent integral leg, then each triangle cannot have an

integral hypotenuse unless they are identical; thus, if that shared leg were part of an integral

cuboid, then it cannot produce an integral end-face diagonal, an integral lateral-face diagonal,

and also an integral main-cube diagonal all at the same time...

...

the best possible answer would be a cuboid with edges 3, 4, an end-face diagonal of 5, a

cube edge of 12, and a main diagonal of 13.

...

this has been proven by method of exhaustion!

(OR) you can watch an endless video of all the calculated 'near misses'. what an absolute waste

of time!

*QED

...

a short theorem: if 2E >= 3V, then Hadwiger's conjecture is true:: proof: assume 2E < 3V; if

we let V= 4, then E < 6, and the complete graph, K4, could have no more than 5 edges and

wouldn't require 4 differently-colored vertices. if 2E >= 3V, then using Euler's characteristic,

2V -2E +2F = 4, and by substitution 2F -V >= 4; also, 6F -3V >= 12, -2E < -3V, & 3F -E >= 6;

thus, {3F -E >= 6} minus {2F -V >= 4} implies that V -E +F >= 2; thus, if the graph is "at least"

planar, V >=4, and 2E >=3V, then Hadwiger's conjecture is absolutely valid. of course, we

are assuming that the graph is simply connected; the one {4-vertex, 6-edge} planar graph is

K4, the two non-planar graphs are both a square with 2 diagonals that cross but do not form

a vertex and a solid tetrahedron! or... we'd have some multiple of these three cases!

*QED 03/4/2012

...

...

I invited a mathematcian to view the next two proofs, and he said that they were not rigorous?

I told him that they contained equations that were only manipulated using algebra that an 8th-

grader could understand, and then I asked him to stop contacting me; obviously, he couldn't

read... he still contacted me and could not elaborate on why he didn't not like my derivations.

I attended a lecture by Dr. Paul Halmos (who died in 2006); he insisted that we fight the math

problems... not the eventual answers!

...

the minimum crossing number, cr(K[m,n]) of a complete bipartite graph by me for Zarankiewicz

as Cauchy did for Euler's topological characteristic formula, V -E +F = 2.

...

cr(K[m,n])= floor(m/2)*floor((m-1)/2)*floor(n/2)*floor((n-1)/2)

...

proof:

draw n (even) points in two parallel rows; draw all possible lines from each point to the rest.

divide by 2 to remove the duplicate connections with regard to orientation and divide by 2

again to count only those lines from a point in one row to the others excluding the point di-

rectly opposite the ones in question. thus, we have K(hat)[n,n]= floor(n/2)*floor((n-1)/2) con-

nections; just imagine that we don't connect the points that are immediately adjacent.

...

now, let 'm' equal the number of points in the top row and 'n' equal the number of points in the

bottom row. since two lines and only two lines are allowed to produce a crossing, we can sub-

stitute floor(m*n)/2 in for each 'n' in the 'connection-counting' transformation.

this results in two functions that describe the number of crossings for a set of points in two par-

allel rows. thus, cr(K[m,n])= min{floor(m/2)*floor(n/2)*floor((m*n-1)/4)}= min{floor(m/2)*floor(n/2)

*floor((m+1)/2)*floor((n-1)/2), floor(m/2)*floor(n/2)*floor((m-1)/2)*floor((n+1)/2)}= floor(m/2)*floor

((m-1)/2)*floor(n/2)*floor((n-1)/2) since (m*n-1) is actually the difference of two squares where 'm'

and 'n' are half of the original 'n' points.

...

thus, we have cr(K[m,n])= floor(m/2)*floor((m-1)/2)*floor(n/2)*floor((n-1)/2) as the supremum or

unique minimum of the topological object in question.

...

therefore, cr(K[m,n])= floor(m/2)*floor((m-1)/2)*floor(n/2)*floor((n-1)/2).

*QED

...

...

the minimum crossing number, cr(K[n,n]) of a complete graph by me for Richard K. Guy who's 94

years old at the rendering date of this proof to him. initially, he and his protege said that they

weren't interested in the proof, but I noticed that someone from Calgary visited my website on the

very next day... yeah, right...

...

cr(K[n,n])= (1/4)*floor(n/2)*floor((n-1)/2)*floor((n-2)/2)*floor((n-3)/2)

...

proof:

use K(hat)[n,n]= floor(n/2)*floor((n-1)/2) to count the number connections regardless of orientation

and considering that points must be placed in 2 concentric circles to achieve the desired result. re-

place 'n' with 'floor(n*(n-2))/4' to count the total number of crossings in a complete graph, spanning

the set from selected points; thus, cr(K[n,n])=min{(1/4)*floor(n/2)*floor((n-2)/2)*floor((n*(n-2)-1)/4)}=

min{floor(n/4)*floor((n-2)/4)*floor(((n-1)*(n-1))/4), floor(n/4)*floor((n-2)/4)*floor((n-3)*(n-1))/4)}=(1/4)*

floor[(n/2)*((n-1)/2)*floor((n-2)/2)*floor((n-3)/2)] where the polynomial 'n*(n-2)-1' can be manipulated

since it's closely divided by 4 within its own floor function. hence, the supremum becomes...

...

cr(K[n,n])= (1/4)*floor(n/2)*floor((n-1)/2)*floor((n-2)/2)*floor((n-3)/2).

*QED

...

NOTE: fewer partitions would mean that we couldn't count using the crossing-transform because it

wouldn't exist, and additional partitions would increase the number of multiplied, floor functions!

the arguments that I'm receiving about the minimum crossings' proofs is that I'm only coming from

above and not from below. that isn't true. I'm coming from above by using the transform floor(n/2)*

floor((n-1)/2) to produce "average counting functions" and selecting a supremum which is unique

by its very definition. the only way to come from below is to divide by '3' or some larger number

in either the transform or the spanning portions, and that would give you the wrong counts!

...

it's very, very difficult to believe that either Zarankiewicz or Guy could have come up with either of

the supremums of two of the minimum counting functions without prior knowledge of the functions

themselves; it's definitely not a simple question or answer, but both persons just happened to pick

the 'needle-out-of-a-haystack' for answers from a long list of careful observations from an even long-

er list of articulated drawings; that's too much work!

...

...

Proth's theorem extended (not by much, but w/out sacrifice!):

...

let Q= k*2^n +1, n >2 is a natural number and k <= 2^n+1. If for some 'a', a^((Q-1)/4) == +/-1 (mod Q),

then 'Q' is prime. (this theorem lays hidden unless also gcd(n, 3) = 1.)

...

proof:

if 'm' is from the set of natural numbers, then every odd prime divisor 'q' of a^(2^(m+1))+/-1 implies

that q == +/-1(mod a^(m+2)) [concluded from generalized Fermat-number 'proofs' by Proth and ad-

justed by me replacing 'm' with 'm+1'].

...

now, if 'p' is any prime divisor of 'R', then a^((Q-1)/4) = (a^k)^(2^(n-2)) == +/-1(mod p) implies that p

== +/-1 (mod 2^n).

...

thus, if 'R' is composite, 'R' will be the product of at least two primes each of which may have a maxi-

mum value of (2^n +1), and it follows that...

...

k*2^n +1 >= (2^n +1)*(2^n +1) = (2^n)*(2^n) + 2*(2^n) +1; but the 1's cancel, so k*(2^n) >= (2^n)*(2^n)

+2*(2^n) and upon dividing by 2^n... implies that k>= 2^n +2.

...

hence, for some 'a', if k <= 2^n +1 and a^((Q-1)/4) == +/-1 (mod Q), then 'Q' is prime.

*QED

...

this simple improvement isn't significant enough to be published, but since I was able to squeeze a

little more out of his 1878-proof, it became the idea that persuaded me to look at other proofs.

...

...

(my) Riesel's theorem (as it should be named):

...

let R= k*2^n -1, n >2 is a natural number and k <= 2^n -1. If for some 'b', b^((R-1)/2) == +/-1 (mod R),

then 'R' is prime; (similar to Proth's theorem, and this theorem lays hidden unless also gcd(n, 3) = 1.)

...

proof:

if 'm' is from the set of natural numbers, then every odd prime divisor 'r' of a^(2^m) +1 implies that q

== +/-1(mod a^(m+1)) [concluded from generalized Fermat-number 'proofs' by Proth, and after my ex-

amination of Proth's theorem].

...

now, if 'q' is any prime divisor of 'S', then b^((R-1)/2)= (b^k)^(2^(n-1)) == +1 (mod q) implies that q

== +/-1 (mod 2^n).

...

thus, if 'S' is composite, 'S' will be the product of at least two primes each of which may have a maxi-

mum value of (2^n -1), and it follows that...

...

k*2^n +1 = (2^n)*(2^n) -1 = (2^n +1)*(2^n -1) <= (2^n)*(2^n) -2*(2^n) +1; implies that -2 <= -2*(2^n)

and dividing by 2^n and multiplying by -1, we have.. 1 > 2^n, contradiction!

...

hence, for some 'b', if k <= 2^n -1 and b^((R-1)/2) == +/-1 (mod R), then 'R' is prime.

*QED

...

Note: a Mersenne number can be easily checked for primality using only one Riesel theorem test!

...

...

some Mersenne divisors theorems:

(all of them can be proved similar to this one)

...

I. theorem: if k >1 and p= 4k +1 is 'prime', then 2*p +1 is also 'prime' iff 2*p +1 divides Mp +2;

a 'complimentary' proof to that of Euler's result for mersenne divisors [shown by modifying La-

grange's proof of the original if p= 4k +3,...].

...

proof(forward): suppose that q= 2*p +1 is prime; q= 3 (mod 8) and 2 is a quadratic residue

modulo q, and there exists an integer 'n' such that n^2= 4 (mod q). Thus, 2^p = 2^((q-1)/2) =

n^(q-1) = -1 (mod q); adding 2 to each portion of this equation shows that (2*p +1) | (Mp +2).

...

proof(backward): let 2*p +1 be a composite factor of Mp +2 & let 'q' be the least prime factor;

2^p= 1 (mod q) and the order of 2 divides both p and q-1; so, the conclusion is the same as

the original Lagrange proof, (2*p +1) +1 > q^2 > p^2 contradicts that p> 2.

*QED

...

II. if p= 4k +3 is 'prime', then 2*p +1 is also 'prime' iff 2*p +1 divides Mp, or 2^p== 1 mod (2p +1).

...

III. if k >1 and p= 4k +1 is 'prime', then 2*p +3 is also 'prime' iff 2*p +3 divides Mp -p, or 2^p -p==1

mod (2p +3).

...

IV. if p= 4k +3 is 'prime', then 2*p +3 is also 'prime' iff 2*p +3 divides Mp -p -1, or 2^p -p == 2

mod (2p +3).

...

...

this proof needs to be corrected. I'm in the process of doing so...

...

Olsen's conjecture is actually a theorem. his conjecture is equivalent to the following as posted

on the Open Problem Garden by M. DeVos. for every finite multiplicative group G, let z(G) denote

the smallest integer 'm' such that every sequence of 'm' elements of G has a sub-sequence of length

|G| with a product equal to 1 in the given order (so Olson's conjecture is equiv. to z(G) <= 2*|G| -1).

it is clear that z(G) <= |G|(|G| -1) +1, since any sequence of length > |G|(|G| -1) must contain at least

|G| copies of the same element, and the product of these will be 1.

...

this statement is equivalent to... if... z(G) <= |G|(|G| -1) +1, then... z(G) <= 2*|G| -1.

it's very true... no one just put pencil to paper; it can be found on the Open Problem Garden!

...

proof: assume that #1.) z(G) > 2|G| -1; we know comfortably that #2.) z(G) <= |G|(|G| -1) +1, and

solving for #1.), we have... [z(G) +1]/2 > |G|. so, by subst. of #1.) into #2.), #2.) becomes z(G) <=

([z(G) +1]/2)*(([z(G) +1]/2) -1) +1. so, 4*z(G) <= (z(G))^2 +2*z(G) +1 -2*z(G) -2 +1 <= (z(G))^2,

or... 4*z(G) <= (z(G))^2... and... z(G) >= 4. and since #1.)... 5/2 >= |G| implies that |G| = 1 or |G| =

2, but by #2.), either z(G) <= 1 or z(G) <= 3; both equalities from #1). are contradictory to #2.),

z(G) >= 4. hence, by contradiction... z(G) <= 2|G| -1.

*QED

01/29/2013

...

...

finally,...

a perfect cuboid cannot exist; I don't know why this problem has presumably existed for so

long; here are the closest solutions...

...

honorable mention:

...

if two non-planar rectangles share an adjacent integral edge and both have integral diagonals,

then if that shared edge were to be a part of an integral cuboid, then those rectangular faces

cannot produce an integral main-cube diagonal or an integral diagonal for the remaining lateral-

or end-face, however you see it...

...

the stereo-typical result would be a cuboid with edges 3, 4, and 4.

...

runner-up:

...

if two non-planar triangles share an adjacent integral leg, then each triangle cannot have an

integral hypotenuse unless they are identical; thus, if that shared leg were part of an integral

cuboid, then it cannot produce an integral end-face diagonal, an integral lateral-face diagonal,

and also an integral main-cube diagonal all at the same time...

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the best possible answer would be a cuboid with edges 3, 4, an end-face diagonal of 5, a

cube edge of 12, and a main diagonal of 13.

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this has been proven by method of exhaustion!

(OR) you can watch an endless video of all the calculated 'near misses'. what an absolute waste

of time!

*QED

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