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1. Consider the smaller numbers Y and Z first. Let Y+Z=a^2, Y-Z=b^2.
Z is either twice a difference of squares, or twice a difference of quarter-squares , but in the latter case, if Z is an integer then it is even. Let Z = 2z.
Since we can consider X and Y in just the same way, Y must also be even, say 2y.
Now:
1.Let (X-Z)=A^2, let (X+Z)=B^2
2.Let (Y-Z)=C^2, let (Y+Z)=D^2
3.Let (X-Y)=F^2, let (X+Y)=G^2
From 2, 2Y=4y=C^2+D^2, so C and D are both even.
Since C is even, C^2 is the even component of two triples: (A^2, C^2, F^2) and (G^2, C^2, B^2), because 2X=A^2+B^2=F^2+G^2, and A^2-F^2=G^2-B^2 = C^2. But non-trivial solutions to this arrangement are not possible:
Proof
Using the Pythagorean identity, assume that C=2uvw. With this C, we can make the 6 triples:
u^2-(vw)^2, 2uvw, u^2+(vw)^2
(vw)^2-u^2, 2uvw, (vw)^2+u^2
v^2-(uw)^2, 2uvw, v^2+(uw)^2
(uw)^2-v^2, 2uvw, (uw)^2+v^2
(uv)^2-w^2, 2uvw, (uv)^2+(w)^2
w^2-(uv)^2, 2uvw, w^2+(uv)^2.
None of these 'work'.
Say, for example, that we set G = (uv)^2+w^2, B = (uw)^2+v^2, F = (uv)^2-w^2, and A = (uw)^2-v^2. Since X = 1/2(A^2+B^2), and X =1/2(F^2+G^2):
I X = 1/2(((uw)^2-v^2)^2+((uw)^2+v^2)^2), and
II X =1/2(((uv)^2-w^2)^2+((uv)^2+w^2)^2), but then, simplifying,
III X=(uw)^4+v^4, X=(uv)^4+w^4, with no non-trivial solutions.
The same method can be repeated, if desired, with every possible permutation of {u,v,w} to complete the proof.
2. Starting with:
I (X-Z)^(1/2) =A, (X+Z)^(1/2) = B, then
II ((X-Z)^(1/2)*(X+Z)^(1/2))^2=(AB)^2=X^2-Z^2, with X the hypotenuse of a Pythagorean triple.
Similarly:
I (X-Y)^(1/2) =F, (X+Y)^(1/2) = G, when
II (FG)^2+Y^2=X^2
Now as before we have 2Z= (A^2-B^2), 2Y=(G^2-F^2), where Z is even and Y is even, so for odd X (since even X will characteristically produce a larger solution), {A,B,F,G} are all odd. Let F=(2f+1), G=(2g+1).
Then:
I 2X=(2f+1)^2+(2g+1)^2, X^2=((2f+1)(2g+1))^2+4y^2, so that
II y = g*(g+1)-f*(f+1), and
III X=2f(f+1)+2g(g+1)+1.
Further:
IV 2X=(2a+1)^2+(2b+1)^2, X^2=((2a+1)(2b+1))^2+4z^2, when
V z = b(b+1)-a(a+1), and
VI X=2a(a+1)+2b(b+1)+1
However, we are interested in cases where the same X satisfies the equation in alternate ways : one possible method is to search for multiple solutions to 2k1(k1+1)+2k2(k2+1)+1=X using Excel.
Solutions with X<1000 are:
{65,125,185,205,305,325,377,425,445,485,493,505,533,565,585,625,629, 689,697,725,745,765,785,793,845,901,905,925,949,965,985}.
Of these, only {125,533,697} produce a square in either (Y+Z) or (Y-Z):
So the minimum solution with X+Y, X-Y, X+Z, X-Z, and Y+Z all square with {X,Y,Z} positive and distinct is {125,100,44}, and the minimum solution with X+Y, X-Y, X+Z, X-Z, and Y-Z all square with {X,Y,Z} positive and distinct is {697,672,528}. |
