12.3 Form: a+b±c = n(d+e±f)

 

Piezas

 

It turns out the system,

 

a+b±c = n(d+e±f)          (eq.1)

a²+b²+c² = d²+e²+f²     (eq.2)

a⁴+b⁴+c⁴ = d⁴+e⁴+f⁴     (eq.3)

 

also has a soln for any n of which Sinha’s problem was just the special case n = 2.  The general problem is equivalent to finding rational points on a certain elliptic curve (though this curve may be degenerate for some n).  To show this, first express the variables in the form,

 

{a,b,c,d,e,f} = {p+q, r+s, t+u, t-u, r-s,  p-q}

 

reducing the degrees of eq.2 and 3, a trick which will be also useful when dealing with sixth powers.  As the explicit expressions are a bit messy, only the steps will be given:

 

1. Solve for p,u using eqs.1,2. 
2. Substituting these into eq.3, it has a quadratic factor in r with discriminant D.
3. D is a quartic in s.  Hence this must be made a square with two small solns for any n as s = q±t.  By treating D = y² as an elliptic curve, from an initial soln subsequent ones can then be found.

Example: Using Sinha’s problem n = 2, and the negative case a+b-c = 2(d+e-f), doing the substitutions, we get,

 

3p-q-r+3q-3t+u = 0                  (eq.1)

pq+rs+tu = 0                             (eq.2)

p³q+pq³+r³s+rs³+t³u+tu³ = 0    (eq.3)

 

Solve for {p,u} and substitute into eq.3.  One factor will be a quadratic in r.  The square-free discriminant D of this is of form,

 

D: = c₁s⁴+c₂s³+c₃s³+c₄s+c₅² = y²

 

with a square constant term and where the c_i are polynomials in {q,t}.  Four small solns are:  s = {q, q-t, q+t, (q-t)/7}. 

 

These may then be used to generate subsequent ones.  For general n, using s = q+t with t = 1, a quadratic form soln can be found for the + case of eq.1 as,

 

{a,b,c} = {1+2nq+r,  n(1+2q),  -1+n-r}

{d,e,f} = {n+q+nq+r,  1-n+q-nq-r,  1+2q}

 

where r = q+(n+1)q², for arbitrary {n,q}.  This also satisfies {a-b+c, d+e-f} = {0,0} though one can also find parametric ones such that this is not the case.

 

 

12.4 Form: na+b+c = d+e+nf

 

Piezas

 

Just like the previous, this has a soln for any n.  The inspiration for this form is the smallest polynomial soln to the sixth power multi-grade a^k+b^k+c^k = d^k+e^k+f^k,  k = 2,6 which, among others, surprisingly obeys the side condition 3a+b+c = d+e+3f.  (Other n for sixth powers obeying na+b+c = d+e+nf are now known.)  For the quartic version, for general n again this can be reduced to finding rational points on an elliptic curve.  Let,

 

na+b+c = d+e+nf          (eq.1)

a²+b²+c² = d²+e²+f²     (eq.2)

a⁴+b⁴+c⁴ = d⁴+e⁴+f⁴     (eq.3)

 

and likewise {a,b,c,d,e,f} = {p+q, r+s, t+u, t-u, r-s,  p-q},

 

1. Solve for r,u in eq.1,2. When substituted into eq.3, this becomes a quadratic in q.
2. Discriminant D of eq.3 is a quartic polynomial in s (but a sextic in p,t).
3. For general n, four small solns are s = {p-nt,  p-t,  p+t,  p}.  Using these and others, subsequent ones can then be found.  Thus, any soln to eq.1,2,3 must solve D = y².

 Example: Using s = p+t with t = 1 yields,

 

{a,b,c} = {2n-3p+np,  -n-n²+3p+np,  -n(-1+n-2p)}

{d,e,f} = {n(1+n-2p),  -n+n²+3p-np,  -2n+3p+np}

 

for arbitrary {n,p}, with this particular soln also satisfying {a+b-c, d-e+f} = {0,0}.

 

 

12.5 Form:  na+b = e+nf

 

Piezas

 

For n = 1, this can be seen as the case n = 0 of a+b+nc = nd+e+f of the system above, and can be completely solved as polynomials thus providing a four-parameter soln to a^k+b^k+c^k = d^k+e^k+f^k, for k = 2,4.  Interestingly, this ultimately involves solving the simple equation x²+my² = z².  Furthermore, any soln implies a^k+b^k+c^k+d^k = e^k+f^k+g^k+h^k for k = 1,2,3,5.

 

Theorem:  If (p+u)^k + (p-u)^k + (2q)^k = (p+v)^k + (p-v)^k + (2r)^k,  for k = 2,4, then,

 

(p+2q)^k + (p-2q)^k + (-p+2r)^k + (-p-2r)^k = (-2p+u)^k + (-2p-u)^k + (2p+v)^k + (2p-v)^k,  for k = 1,2,3,5

 

The complete soln is given by the simultaneous eqns,

 

-3p²+q²+3r² = u²

-3p²+3q²+r² = v²

 

Proof:  Just solve for {u,v} in terms of {p,q,r}.  This is simpler than it looks since the complete parametrization to both is in linear forms.  Alternatively, one can just directly solve the form,

 

a^k+b^k+c^k = d^k+e^k+f^k for k = 2,4 (or Q₁)

 

with a+b = d+e, eliminate e,f using resultants, giving the final eqn,

 

2a²+5ab+2b²-2c²-3ad-3bd+3d² = 0

 

Solving this for a, its discriminant must be made a square and after suitable transformations, turns out to involve the eqn,

 

x₁²+nx₂² =  x₃²+nx₄²

 

which has a complete soln.  Sparing the reader some algebra, the final eqn, with e = a+b-d,  is solved as,

 

{a,b,c,d,e} = {-3pr-4qr+2ps+qs,  5qr-ps,  3pr+qs,  3qr+ps, -3pr-2qr+qs}

 

It remains to find f.  Substituting these values into Q₁, one ends up with a quadratic function to be made a square.  To simplify matters, let,

 

{f, r,s} = {(4p+2q)z,  (2p+q)²y,  2x+(6p+q)(p+2q)y}

 

and this final condition becomes,

 

x²-6q(3p+q)(p²-q²)y² = z²

 

This is an eqn of form x²+ny² = z² which can be completely solved as {x,y,z} = {u²-nv²,  2uv,  u²+nv²} so we now have a four-parameter soln in the variables {p,q,u,v}.  For a specific example, let {p,q} = {0,1} and we get the nice identity,

 

(5y)^k + (x-2y)^k + (x+2y)^k = x^k + (3y)^k + z^k, 

 

if x²+24y² = z².  (And so on for all p,q.)   An alternative formulation can be given as,

 

Theorem:  If (p-q+r)^k + (p+q-r)^k + x^k = (p+q+r)^k + (p-q-r)^k + y^k,  for k = 2,4, then,

 

(2p+q+r)^k + (2p-q-r)^k + (-2p-q+r)^k + (-2p+q-r)^k = (p+x)^k + (p-x)^k + (-p+y)^k + (-p-y)^k,  k = 1,2,3,5

 

Note that the k = 2,4 system still has x₁+x₂ = y₁+y₂.  The complete soln is given by,

 

{r,p} = {(x²-y²)/(8q),  s/(24q)}, where s² = -192q⁴ +96(x²+y²)q²-3(x²-y²)²

 

so the problem is to find {q,x,y} which makes the quartic a square.  However, we already know that this variant of Q₁ has a polynomial soln so after a little reverse algebra, we find,

 

{q,s,x,y} = {vw/2, 6vw(2t+vw), uw+v, uw-v},  with w = (3tv+z)/(2u²-2v²),  if 4(u²-v²)² + 3(4u²-v²)t² = z²

 

Just like before, the last condition is of the form x²+dy² = z² and is easily solved completely as,

 

t = 4mn(u²-v²)/(m²-3(4u²-v²)n²)

 

for free variables {m,n,u,v}.  Given the constraint na+b = e+nf on Q₁, we have discussed the case n = 1.  For general n,

 

na+b = e+nf                   (eq.1)

a²+b²+c² = d²+e²+f²     (eq.2)

a⁴+b⁴+c⁴ = d⁴+e⁴+f⁴     (eq.3)

nb+c = -(nd+f)               (eq.4)

 

where again {a,b,c,d,e,f} = {p+q, r+s, t+u, t-u, r-s,  p-q} and the fourth eqn is optional.  Solve for {u,s} in eq.1,2.  Substituted into eq.3 this becomes the quadratic in q, 

 

v₁q² = v₂t²,  where v₁ = (p-nr)³-(p-n³r)t²,  v₂ = (p³-nr³)-(p-nr)t² 

 

Thus the objective is to find v₁v₂ = y² which is only a quartic in t.  Four small solns are:  t = {p+r,  p-r,  (p-nr)/(n+1),  (p-nr)/(n-1)}.

 

Non-trivial soln can be computed from these trivial ones.  If the fourth eqn is also to be used, solve for q and eq.3 becomes a quadratic in t with a discriminant D that is only a quartic in p,r.  Without loss of generality, let r =1 and this is,

 

D:= (4+n²)²p⁴ - 4n(4+5n²)p³ + 2(4+3n²+4n⁴)p² - 4n(5+4n²)p + (1+4n²)²

 

Since this has a square leading and constant term, this is easily solved with three small solns:  p = {n,  1/n,  -1/n}.

 

 

12.6 Form:  a+d = n(c+f)

 

Lastly, we have the system,

 

a+d = n(c+f)                  (eq.1)

a²+b²+c² = d²+e²+f²     (eq.2)

a⁴+b⁴+c⁴ = d⁴+e⁴+f⁴     (eq.3)

 

Let {a,b,c,d,e,f} = {p+q, r+s, t+u, t-u, r-s,  p-q}.  Solve for u,p.  Substituted into eq.3, this has a quadratic factor in r.  The square-free discriminant D of this is a quartic in s.  Small solns to finding rational points on the curve D = y² are:  s = {q+t,  q-t,  (1+n)q+(1-n)t}.

 

Note 1:  Any other system consisting of Q₁ and one linear relation between the terms (or two, as long as both are new) which would suffice to reduce the problem to solving a quadratic or an elliptic curve?

Note 2:  There is another system but it involves a quadratic relation between the terms,

 

J. Chernick

 

If pq(p²-q²) = rs(r²-s²), then (p²+q²)^k + (r²-s²)^k + (2rs)^k = (2pq)^k + (r²+s²)^k + (p²-q²)^k,  for k = 2,4.

 

This can be re-phrased as,

 

If a²+b² = c², and d²+e² = f², and ab = de, then a^k+b^k+c^k = d^k+e^k+f^k,  for k = 2,4.  A soln was given by Gloden as,

 

{a,b,c,d,e,f} = {p²-q², p²+r², 2pq, p²+q², p²-r², 2pr},  where p² = q²+qr+r², and can be solved by {p,q,r} = {u²+uv+v²,  u²-v², 2uv+v²}.

 

This form is useful for solving a special case of Descartes' Circle Theorem when all variables are squares,

 

2(w⁴+x⁴+y⁴+z⁴) = (w²+x²+y²+z²)²

 

discussed in the next page, Form 18.

 

 

13. Form: a^k+b^k+c^k = 2d^k+e^k  (k = 2,4)

 

The case e = 0 was studied by Ramanujan.  To recall, he gave the formulas for k = 2,4,

 

a^k + b^k + c^k = 2(ab+ac+bc)^k/2

 

a^k(b-c)^k + b^k(a-c)^k + c^k(a-b)^k = 2(ab+ac+bc)^2k/2

 

(a²b+b²c+c²a)^k + (ab²+bc²+ca²)^k + (3abc)^k = 2(ab+ac+bc)^3k/2

 

(and so on) with a+b+c = 0.  As was already mentioned, this form gives rise to a fifth-power multi-grade with just 10 terms,

 

(a+d)^k + (-a+d)^k + (b+d)^k + (-b+d)^k + (c+d)^k + (-c+d)^k = (-2d)^k + (-2d)^k + (e+d)^k + (-e+d)^k,  k = 1,2,3,4,5. 

 

For the complete radical soln to the system x₁^k+x₂^k+x₃^k = 2x₄^k+ x₅^k, k = 2,4, express it in the form,

 

(2a)^k + (2b)^k + (p+q)^k = 2(2c)^k + (p-q)^k

 

which gives,

 

a²+b²-2c²+pq = 0

2a⁴+2b⁴-4c⁴+p³q+pq³ = 0

 

a useful trick to reduce the degree.  Combining the two by eliminating q then gives the beautifully simple condition,

 

(a²+b²-2c²)p⁴ - 2(a⁴+b⁴-2c⁴)p² + (a²+b²-2c²)³ = 0

 

Unfortunately, the discriminant of this is a sextic in a,b (and an octic in c) so cannot be treated as an elliptic curve.  However, it does have solns one of which, implied in Ramanujan’s case when last term is p-q = 0, is

 

{a,b,c,p} = {u²-v²,  2uv+v²,  u²+uv+v²,  u²+2uv}

 

which makes the discriminant zero.  (There are other solns as well.)

 

Piezas

 

Parametric solns to a^k+b^k+c^k = 2d^k+e^k have been found in two cases: 

 

I.  When a+b = c. 

1. If e = 0, then a²+ab+b² = d².  (Ramanujan’s)
2. If e = 2d, then a²+ab+b² = 3d².

II.  When a+b = 2d. 

 

Let, (p+r)^k + (-p+r)^k + q^k = 2r^k + s^k,  then p²+2q² = 6r², and 2p²+q² = s². 

 

The first eqn can be solved as {p,q,r} = {2u²-4uv-4v²,  u²+4uv-2v²,  u²+2v²} and substituting this into the second gives the elliptic curve,

 

(3u²-4uv-6v²)² + 32u²v² = s²

 

with one non-trivial soln being {u,v} = {12, 5}, and so on.  Note: Any other case?

 

 

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