025: Sixth Powers (8 terms)

 
 

(Update, 1/25/10):   Piezas

 

In the previous section, it was shown how a k = 1,3,5 can lead to a k = 2,4,6.  This can be brought higher.  Given a (k.4.4) for k = 2,4,6, if its terms {xi , yi} are such that x1-x2 = y1-y2 = y2-y3 , then it can lead to a (k.5.5) for k = 1,3,5,7.  This is exemplified by the infinite family,

 

(3a+2b+c)k + (3a+2b-c)k + (-2a-b+d)k + (-2a-b-d)k = (3a+b+2c)k + (3a+b)k + (3a+b-2c)k + (-7a-b)k,  for k = 1,2,4,6

 

which leads to the higher system,

 

[3a+b+3c, 3a+2b-2c, -7a-b+c, -2a-b-c-d,  -2a-b-c+d]k =

[3a+b-3c, 3a+2b+2c, -7a-b-c, -2a-b+c-d,  -2a-b+c+d]k,  for k = 1,2,3,5,7

 

where, for both, 10a2-b2 = c2, and 55a2-6b2 = d2, with trivial ratios a/b = {1, 1/3, -5/13}, and small non-trivial soln {a,b} = {5, 13}.  It is tempting to speculate that there may be higher-order version of this for (k.5.5), k = 2,4,6,8, with the form,

 

[pa+qb+c, pa+qb-c, ra+sb+c, ra+sb-c, ta+wb]k =

[ua+vb+2c, ua+vb, ua+vb-2c, xa+yb+d, xa+yb-d]k   (System 3)

 

where m1a2+m2b2 = c2, and m3a2+m4b2 = d2, for constants {p,q,r,s,t,u,v,w,x,y} and {m1, m2, m3, m4}.  System 3 has x1-x2 = x3-x4 = y1-y2 = y2-y3 and would automatically lead to a (k.6.6) for k = 1,3,5,7,9.  While there is numerical example with the right constraints, it is unknown if it belongs to a family with an infinite number of rational solns.  (End update.)

 

Crussol

 

Given the system of eqns for indefinite k,

 

x1k + x2k + x3k + x4k = y1k + y2k + y3k + y4k

 

Chernick used certain expressions to solve the case k = 1,2,3 while Lander focused on k = 1,3,5, (which Choudhry showed could be extended to k = 1,3,7).  But prior to this, Crussol took the case k = 1,2,4,6 and used the version,

 

(a+x)k + (a-x)k + (b+y)k + (b-y)k = (a+z)k + (a-z)k + (b+t)k + (b-t)k        (eq.0)

 

which in one sense is a simpler form and gave a partial soln.  (Later we will give the complete one.)  Crussol established that for k = 2,4,6, one should satisfy the beautifully simple conditions,

 

x2+y2 = z2+t2

5(a2+b2) = x2+y2

3(a2-b2) = y2-z2
 

The terms of eq.0, expressed as {xi, yi}, obviously is identically true for the side conditions given by Lander, namely x1+x2 = y1+y2,  x3+x4 = y3+y4 which makes it already valid for k = 1.  However, if also true for k = 2,4,6, then it satisfies the following additional relations,

 

x1x2+x3x4 = y1y2+y3y4

 

x12+3x1x2+x22+x32+3x3x4+x42 = 0

 

y12+3y1y2+y22+y32+3y3y4+y42 = 0

 
To solve the first condition, Crussol used {x,y,z,t} = {de+cf,  ce-df,  de-cf,  ce+df} which reduced the other two to,

 

5(a2+b2) = (c2+d2)(e2+f2)          (eq.1)

3(a2-b2) = (c2-d2)(e2-f2)            (eq.2)

 

Note that eq.1 is of the form m(a2+b2) = (c2+d2)(e2+f2) for some constant m and can be solved, though not completely (see section on Sums of Two Squares), as,

 

{a,b,c,d} = {fu-ev,  eu+fv,  2u+v,  -u+2v}

 

for free variables {e,f,u,v}.  Using these expressions on (eq.2) results in a quadratic in e (or f) and solving for this gives the last two unknowns,

 

{e,f} = {3u2+4uv-3v2,  3uv±w}

 

where {u,v,w} satisfy the quartic elliptic curve,

 

(3u2+4uv-3v2)2 + (3uv)2 = w2

 

two small solns of which are {u,v} = {5,3} or {9,16}, etc. 

 

Note 1:  For comparison, when dealing with odd powers, one can move terms around and it makes no difference to use the related system,

 

(a+x)k + (a-x)k + (-a+z)k + (-a-z)k = (b+y)k + (b-y)k + (-b+t)k + (-b-t)k

 

where now it has x1+x2+x3+x4 = y1+y2+y3+y= 0.  But for even powers, this only has trivial solns if simultaneous true for k = 2,4,6.

 

Note 2:  The basic form of eq.0 can also be used for just k = 1,2,6.  For example, Shuwen found,

 

31k + 126k + 62k + 107k = 38k + 119k + 51k + 118k

 

which has 31+126 = 38+119, 62+107 = 51+118 and apparently is a particular instance of a parametric identity though I haven’t been able to derive it yet. (Anyone can provide it?)  Whether Crussol’s form can be extended to eighth powers for k = 1,8 or 1,2,8 still remains to be seen.  (The only known numerical soln to (8.4.4) found by Kuosa in 2006 is not of this form.)

 

Note 3:  Using a different approach, the complete soln to k = 1,2,4,6 with x1+x2 = y1+y2, x3+x4 = y3+y4 will be discussed in the subsequent sections.
 
(Update, 1/18/10):  Piezas
 

Crussol’s soln can be given in a more aesthetic manner as two quadratics to be made squares.  Let,

 

[c+be+af, c-be-af, d+ae-bf, d-ae+bf]k =

[c+be-af, c-be+af, d+ae+bf, d-ae-bf]k

 

for k = 1,2,4,6, where,

 

ma2+nb2 = c2     (eq.1)

na2+mb2 = d2     (eq.2)

 

with {m,n} = {(4e2-f2)/15,  (4f2-e2)/15}  for some constants {e,f}.  These simultaneous equations, eq.1 and eq.2, define an elliptic curve.  For ex, let {e,f} = {13, 1} and we get,

 

(a+13b+c)k + (-a-13b+c)k + (13a-b+d)k + (-13a+b+d)k =

(-a+13b+c)k + (a-13b+c)k + (13a+b+d)k + (-13a-b+d)k

 

where 45a2-11b2 = c2, and -11a2+45b2 = d2.  These conditions are the same as the ones for the only known [k.6.6] identity for k = 1,2,4,6,8,10 discussed in Tenth Powers.  More generally, given eq.1 and eq.2, which lead to an elliptic curve E(m,n), it is quite easy to find a k = 1,2,4,6 which uses this curve since the constants {e,f} can be derived from {m,n} as,

 

{4m+n, m+4n} = {e2, f2}

 

hence, if the two expressions themselves are squares, then it can be used for a k = 1,2,4,6.  (End update.)

 

 

(Update, 2/1/10)  This update covers a family that includes 6th, 8th, and 10th power multi-grades.  Whether it can be extended for 12th powers with a minimal number of terms remains to be seen.

 

Piezas  (6th powers)

 

Given the general form,

 

[xy+ax+3by-c,  xy-ax-3by-c,  xy-3bx+ay+c,  xy+3bx-ay+c]k =

[xy+ay+3bx-c,  xy-ay-3bx-c,  xy-3by+ax+c,  xy+3by-ax+c]k

 

Notice how {x,y} just swap places or alternatively, “a” in one side is replaced by “3b” in the other side.  With terms expressed as {pi, qi}, this form identically is true for the following constraints,

 

p1+p2 = q1+q2,

p3+p4 = q3+q4,

p1k+p2k+p3k+p4k = q1k+q2k+q3k+q4k,  k = 1,2

 

We can make this valid for k = 1,2,4,6.  For k = 4, let c = ab.  For k = 6 and some constants {a,b}, then {x,y} must satisfy the simple eqn,

 

10a2b2-(a2+9b2)(x2+y2)+10x2y2 = 0

 

This is an elliptic “curve” in disguise.  Let {x,y} = {u, v/(10u2-a2-9b2)} and one is to solve,

 

(a2+9b2-10u2)(10a2b2-(a2+9b2)u2) = v2

 

which is a quartic polynomial in u to be made a square.  Trivial ratios are a/b = {0,1,3,9}.  It can be proven that for any rational {a,b} that do not have a trivial ratio, one can always find an infinite number of rational {x,y}.  Proof:  A polynomial soln can be given starting from the trivial point u = 3b.   This yields the non-trivial,

 

u = 9b(a2+15b2)/(5a2-117b2)

 

from which one can then successively compute an infinite sequence of rational polynomials. 

 

 

Chris Smyth  (8th powers)

 

This is a variation of the Letac-Sinha 8th power multi-grade.  For k = 1,2,4,6,8,

 

[xy+ax+ay-d, -(xy-ax-ay-d), xy-bx+by+d, -(xy+bx-by+d), cx+cy]k =

[xy+bx+by-d, -(xy-bx-by-d), xy-ax+ay+d, -(xy+ax-ay+d), cx-cy]k

 

where {a,b,c,d} = {1, 3, 4, 11} and {x,y} satisfies the eqn,

 

x2y2-13(x2+y2)+112 = 0

 

One can note its affinity with the 6th power multi-grade and see also in the general form how “a” in one side has just been replaced with “b” in the other side.  With terms expressed as {pi,qi}, this obeys the constraints,

 

p1-p2 = q1-q2

p3-p4 = q3-q4

 

If we set that,

 

p1k+p2k+p3k+p4k+p5k = q1k+q2k+q3k+q4k+q5k,  k = 1,2

 

then one must have 2ak-2bk+ck = 0,  for k = 1,2, and the consequence is that,

 

p1k+p2k+p5k = q1k+q2k, 

p3k+p4k = q3k+q4k+q5k

 

for k = 1,2 which, naturally enough, is the same set of constraints obeyed by the Letac-Sinha identity.  Expanding the system for k = 4,6,8, it will be seen that the only non-trivial soln in the rationals is {a,b,c,d} = {m, 3m, 4m, 11m2} where one can set m = 1 without loss of generality and {x,y} satisfies the eqn,

 

x2y2-13(x2+y2)+112 = 0

 

Let {x,y} = {u, v/(u2-13)} and this is the elliptic curve,

 

(u2-13)(13u2-112) = v2

 

Smyth’s form for (k.5.5) can be generalized by adding two terms on each side to create a (k.7.7).  A possible addition, given in italics, may be,

 

[xy+ax+ay-c, -(xy-ax-ay-c), xy-bx+by+c, -(xy+bx-by+c), dx+dy, ex+fy, fx-ey]k =

[xy+by+bx-c, -(xy-by-bx-c), xy-ay+ax+c, -(xy+ay-ax+c), dx-dy, ey+fx, ex-fy]k

 

which is the same pair successfully added to the (k.4.4).  For non-zero {e,f}, whether this has a non-trivial soln up to k = 8, 10, or even 12 is unknown.

 

 

Choudhry, Wroblewski   (10th powers)

 

The 6th power (k.4.4) identity can be generalized to the form (k.6.6) by adding the italized pair of terms,

 

[-(axy+bx+cy-d), axy-bx-cy-d, axy-cx+by+d, -(axy+cx-by+d), ex+fy, fx-ey]k =

[-(axy+by+cx-d), axy-by-cx-d, axy-cy+bx+d, -(axy+cy-bx+d), ey+fx, ex-fy]k

 

where again, {x,y} merely swap places.  (Since the eqn is homogeneous, one may assume the leading coefficient of the terms as a = 1 without loss of generality.)  This obeys the same basic constraints (up to sign changes),

 

p1-p2 = q1-q2,

p3-p4 = q3-q4,

(-p1)k+p2k+p3k+(-p4)k = (-q1)k +q2k+q3k+(-q4)k,  k = 1,2

p52+p62 = q52+q62

 

If we assume,

 

p1+p2+p5 = q1+q2+q5, 

p3+p4+p6 = q3+q4+q6, 

 

then it must be the case that 2(b-c) = e-f   (eq.1).  Choudhry and Wroblewski found for k = 2,4,6,8,10, the equivalent non-trivial solns,

 

{a,b,c,d,e,f} = {1, 1, 2, 14, 3, 5},  where 2x2y2-17(x2+y2)+392 = 0

{a,b,c,d,e,f} = {2, 1, 2, 7, 3, 5},  where 8x2y2-17(x2+y2)+98 = 0

 

Disregarding eq.1, two other distinct non-trivial solns found (by Piezas), though only up to k = 2,4,6,8, are,

 

{a,b,c,d,e,f} = {5, 3, 7, 3, 2, 10},  where 125x2y2-53(x2+y2)+45 = 0

{a,b,c,d,e,f} = {5, 9, 11, 11, 10, 12}, where 125x2y2-221(x2+y2)+605 = 0

 

though this author is not certain if these are the same 8th power multi-grades found by Wroblewski.  (End update.)

 

 

Piezas

 

To find an infinite family of polynomial solns to k = 1,2,4,6, we will use the form F2,

 

(a+bh)k + (c+dh)k + (e+fh)k + (g+h)k = (a-bh)k + (c-dh)k + (e-fh)k + (g-h)k

 

the big brother of F1 (which was for six terms with k = 2,4 and k = 1,2,6).  To make F2 valid for k = 1,2, set g = -(ab+cd+ef) and f = -(1+b+d).  Expanding for k = 4,6, we get,

 

(Poly11)h2 + (Poly12) = 0                                (eq.1)

(Poly21)h4 + (Poly22)h2 + (Poly23) = 0            (eq.2)

 

respectively, where the Polyi are expressions in the variables vi = {a,b,c,d,e}.  There are two ways to solve these two: first, to make Poly11 = Poly12 = 0, which will also cause eq.2 to factor into two quadratics, or second, to find vi so eq.2 will factor into the same quadratic as eq.1.  To find a common root between eq.1 and 2, it is a simple matter to eliminate the variable h using Mathematica’s Resultant[ ] function. (And since h is conveniently in even powers, it can be set h = √v to shorten the calculation.)  The resultant is a factorable 6th deg in e and, surprisingly, this has three non-trivial linear factors given by,

 

(a+ab-c+cd-be-de) (-a+ab+c+cd-be-de) (a+ab+c+cd-2e-be-de) = 0

 

plus an irreducible cubic factor.  (One can get more solns if a linear root of the cubic can be found such as for special cases like d = -b when the above three become trivial but the cubic now has a single non-trivial linear factor.)  In the general case though, solving for one appropriate variable of either of the three (b,d,e, respectively) and substituting into Poly11 = 0, one can solve for a second variable.  These two can then set Poly11 = Poly12 = 0, and cause eq.2 to factor: two linear and one quadratic in h, with only the latter being non-trivial.  Sparing the reader the rest of the algebra, given F2,

 

(a+bh)k + (c+dh)k + (e+fh)k + (g+h)k = (a-bh)k + (c-dh)k + (e-fh)k + (g-h)k

 

where g = -(ab+cd+ef) and f = -(1+b+d), then for k = 1,2,4,6, if,
 
A. Linear factors:
 
First factor:

 

b = (1+d)(a+2e)/(a-e),  c = (2+d)(a+e)/(1-d), where {a,d,e} satisfy

 

(Poly1)(-1+d)2h2-(Poly2)(a-e)2 = 0

 

Poly1:= a2d-(9+20d+9d2)ae+de2;  Poly2:= 9a2d-(1-20d+d2)ae+9de2

 

For arbitrary {a,d,e,h}, this already solves k = 1,2,4.  For k = 6, the quadratic in h must be solved hence the problem is reduced to making its discriminant D a square, or D = (Poly1)(Poly2) = y2.  (Note that the Polyi are curiously palindromic.)  Since D is a quartic polynomial with a square leading and constant term, this is easily made a square.  For example, using Fermat’s method, one soln is,

 

{a,e} = {9d(41+80d+41d2),  -(5+4d)(4+5d)(20+59d+20d2)}

 

though presumably smaller polynomials may exist.  Using this initial rational point on the curve D = y2, subsequent ones can then be found.

 

Second factor:

 

a = (2+b)(c+e)/(1-b),  d = (1+b)(c+2e)/(c-e), where {b,c,e} satisfy

 

(Poly3)(-1+b)2h2-(Poly4)(c-e)2 = 0

 

Poly3:= bc2-(9+20b+9b2)ce+be2;  Poly4:= 9bc2-(1-20b+b2)ce+9be2

 

so to solve (Poly3)(Poly4) = y2.  (After an exchange of variables this is essentially the same as in the first.)

 

Third factor:

 

b = -(a+2c)(a+c-2e)/(a2-c2+ae-ce),  d = (2a+c)(a+c-2e)/(a2-c2+ae-ce),  where {a,c,e} satisfy

 

(Poly5)h2-(Poly6)(a-c)2(a+c+e)2 = 0

 

Poly5:= (a+c)2(2a+c)(a+2c)+(a+c)(a2-38ac+c2)e-(a2-38ac+c2)e2;  Poly6:= (2a+c)(a+2c)+(a+c)e-e2

 

so to solve (Poly5)(Poly6) = y2.  This again has a square constant term so is easily made a square.
 
B. Alternative method

 

Still more solns can be found by solving Poly11 = Poly12 = 0 alternatively.  Instead of eliminating h between eq.1 and 2, we eliminate the variable e between Poly11 and 12.  This has seven non-trivial factors, with three essentially the same as the ones already discussed.  The other four are the simpler,

 

(2+b+d)(1+b)(1+d)(b+d) = 0

 

The first is a special case of Poly11 = Poly12 = 0, giving a simple soln to F2,

 

(a+bh)k + (c+dh)k + (e+fh)k + (g+h)k = (a-bh)k + (c-dh)k + (e-fh)k + (g-h)k

 

still with g = -(ab+cd+ef) and f = -(1+b+d), where,

 

{c,d,e,h} = {-ab(1-b)/w,  -2-b,  a(1-b)(3+2b)/w,  ay/w},

 

with w = (2+b)(3+b) and b2+2b+21 = y2.

 

This gives terms as quartic polynomials.  (Alternatively, one can set d = 1 to get similar results.)  The next three factors are variations on the same theme, namely a system that also satisfies x1+x2 = y1+y2 and x3+x4 = y3+y4 which imply an F2 with d = -b; f = -1.  Solving Poly11 = Poly12 = 0 gives a soln but is trivial.  However, by using the previous method of eliminating h between eq.1 and eq.2, the resultant gives three factors (squared this time), the first two trivial but not the third.  This, in fact, is the linear root of the irreducible cubic which reduces for the special case d = -b.  Using this root, eq.2 factors into two quadratics, one identical to eq.1 so the problem is reduced to just solving this. 
 
Thus, the complete soln to,

 

 x1k+x2k+x3k+x4k = y1k+y2k+y3k+y4k

 

for k = 1,2,4,6 where x1+x2 = y1+y2 and x3+x4 = y3+y4 is given by F2,

 

(a+bh)k + (c+dh)k + (e+fh)k + (g+h)k = (a-bh)k + (c-dh)k + (e-fh)k + (g-h)k

 

with {d, f} = {-b, -1} where {a,g,h} = {v1(bc+e)e/w,  v2(bc+e)c/w,  y/(2w)}, and w = v1be+v2c,

 

and {b,c,e} satisfying the quartic,

 

(2v2c2-15bce+2v1e2)2 - v1v2c2e2 = y2

 

where {v1, v2} = {b2-4, 1-4b2}.  Since again this has a square leading and constant term, this is easily made a square with one small soln {c,e} = {8b, 1-4b2} giving,

 

{a,g,h} = {(b2-4)(1+4b2)/u,  8b(1+4b2)/u,  (4-13b2+4b4)/u}, 

 

with u = b(b2+4) and yielding terms as 5th degree polynomials in b. From this initial polynomial soln, one can then generate an infinite more. (In contrast to Crussol's approach which can only give infinite numerical solns.)  If we are to consider the more general system,

 

nx1+x2+x3+x4 = ny1+y2+y3+y4

x1k+x2k+x3k+x4k = y1k+y2k+y3k+y4k

 

for k = 2,4,6, using our previous results for k = 1,2,4,6 it can be proven this has a soln for any n.  We simply use the same method of changing the signs of appropriate terms since this does not affect the validity of the even powers.  Thus, the system valid at k = 1,

 

(a+bh) + (c+dh) + (e+fh) + (g+h) = (a-bh) + (c-dh) + (e-fh) + (g-h)

 

where f = -1-b-d, is also good for,

 

(-a-bh) + n(c+dh) + (-e-fh) + (g+h) = (-a+bh) + n(c-dh) + (-e+fh) + (g-h)

 

if d is specialized as d = -2/(n+1).  For example, an explicit soln to (Poly1)(Poly2) = y2 for k = 1,2,4,6 (using the first factor) was given earlier as,

 

{a,e} = {9d(41+80d+41d2),  -(5+4d)(4+5d)(20+59d+20d2)}

 

for any d.  This can be used for k = 2,4,6 with (-x1)+nx2+(-x3)+x4 = (-y1)+ny2+(-y3)+y4, by letting d = -2/(n+1), hence the system can be solved for any n as claimed.  (If n = -1, division by zero can be avoided by simply changing the sign of x2,y2 to change the sign of n.)  Some concise examples for special n, derived slightly differently, are,

 

(a+bh)k + (c+dh)k + (e+fh)k + (g+h)k = (a-bh)k + (c-dh)k + (e-fh)k + (g-h)k

 

where a = 1,  f  = -1-bn-d,  g = -(b+cd+ef), valid for k = 2,4,6 if,

 

1) {b,c,d,e,h} = {-2/(1+n),  -2(-2+n)(3+n)/w,  2/(1+n),  (1+n)(3+n)/w,  (1+n)y/w}, where w = 2n(-5+n) and y2 = 81-2n+n2.

 

2) {b,c,d,e,h} = {-2/(1+n),  -(3+n)(-1+3n)/w,  -1,  -2(3+n)/w,  (1+n)y/w}, where w = 2n(1+3n) and y2 = 21+38n+21n2.

 

3) {b,c,d,e,h} = {-2/(1+n),  -(3+n)(-2+3n)/w,  -n/(1+n),  -(3+n)/w,  (1+n)y/w}, where w = 2n(-1+3n) and y2 = -15+14n+33n2.

 

which also are true for both,

 

n(a+bh) + (c+dh) + (e+fh) + (g+h) = n(a-bh) + (c-dh) + (e-fh) + (g-h)

 

(a+bh) + (-c-dh) + (-e-fh) + (g+h) = (a-bh) + (-c+dh) + (-e+fh) + (g-h)

 

Note 1:  In contrast, as was seen, an infinite number of primitive solns to the system,

 

nx1+x2+x3 = ny1+y2+y3

x1k+x2k+x3k = y1k+y2k+y3k

 

for k = 2,6 and integer n,xi,yi can be proven only for some n which, other than the exceptional case n = 1, for the moment are all n = 3m.  Whether there is an infinite number for some n ≠ 3m remains to be seen.

 

Note 2:  If the terms are signed, all known solns to S6 are valid for k = 1 as well.  (In fact, also for S8 and S10.)  Is there signed S6 that is not for k=1?  A simple computer search may be able to settle this.

 

Note 3:  Lastly, as was mentioned, the system k = 1,2,4,6 with side conditions x1+x2 = y1+y2; x3+x4 = y3+y4 also has solns for just k = 1,2,6.  The complete soln to this would be to ignore eq.1 and just solve eq.2 which, to recall, is a quartic with only even powers.  To factor this into two quadratics entails the sufficient (but not necessary)1 condition of making its discriminant D a square.  As a polynomial in the variable a, D is a quartic with a square leading term so it is quite easy to do this.  A simpler case is when b = -2 since the constant term of D also becomes a square.  Set c = 1 without loss of generality to get,

 

D: = (3-4e+2e2)2 + (-39+88e-64e2+16e3)a + (73-88e+28e2)a2 + 3(-13+8e)a3 + 9a4

 

Three small non-trivial solns to D = y2 are,

 

a1 = 8(2e-e2)/(-17+16e),   a2 = -(21-22e+8e2)/(-19+8e),   a3 = (13-18e+8e2)/(-13+16e),

 

giving three families solving the specialized form of F2 for just k = 1,2,6,

 

(a-2h)k + (1+2h)k + (e-h)k + (g+h)k = (a+2h)k + (1-2h)k + (e+h)k + (g-h)k

 

where g = -2+2a+e and either,

 

1) a = 8(2e-e2)/(-17+16e),  h = y1/(-17+16e), and y12 = (17-23e+8e2)(17+13e+8e2)

 

2) a = -(21-22e+8e2)/(-19+8e),  h = y2/(-19+8e), and y22 = -53+82e-67e2+32e3

 

3) a = (13-18e+8e2)/(-13+8e),  h = y3/(3(-13+8e)), and y32 = 3(2873-7878e+8415e2-4096e3+768e4)

 

Footnote 1:  While it is a reasonable assumption that P: = x4+bx2+c = 0 can be completely factored into two quadratics by finding b,c such that the expression D = b2-4c is a square, a second approach is to find b,c such that it factors as P = (x2+mx+n)(x2-mx+n) = 0 in which case it generally does not have a D that is a square.

 
 
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