2-power teams, and their leaders (AoT, nr.1)


2-power team definition
------------------------------                                    

A 2-power team is a quadruple  A := (a  b  c  d)  of integers such that two conditions are satisfied:

(i)    a  >  b  >/  c >  d  >/  0  --  a team inequality;

(ii)  a^2 + d^2  =  b^2 + c^2 -- a team equation.

Also the two more more equations equivalent to (ii) are called team equations:

(ii')                    a^2 - b^2  =  c^2 - d^2

(ii'')                    a^2 - c^2  =  b^2 - d^2

Then  integer  a  is called the leader of team  A.  We'd like to know everything about the leaders (and about the teams too).

Furthermore, 2-power team is called proper if and only if  d > 0.

Let's have an initial theorem, just for starters; first let's recall that a sum  x+y  and the difference  x-y  of arbitrary integers  x y  have the same parity; now we're ready:

Theorem 0   The sum (as well as the difference) of any two different variables among  a b c d, and the sum (as well as the difference) of the remaining two variables have the same parity.

-----

Note again that the equation

(ii')  a^2 - b^2   =   c^2 - d^2

is equivalent to (ii). Thus, for team  (a b c d)  we obtain:

           c-d  =  (c^2-d^2)/(c+d) = (a^2-b^2) / (c+d) = (a-b)*(a+b) / (c+d)

                  >/  (a-b)*(c+d+2) / (c+d)  >  a-b  >/  1

Thus,

                       c-d  >  a-b

and  c >/ c-d > a-b >/ 1,  i.e.  c > 1  hence

                         c  >/  c-d  >/  2

Remark 0   The following three equations are equivalent:

  • a+d <  b+c
  • a-b  <  c-d
  • a-c  <  b-d
Thus all three hold for arbitrary team  (a b c d).  Now we can state:

(i')     a  >  b  >/  c  >/  d+2  >/  2               for every team  (a b c d)

Thus,
                  d >/ 0         c >/ 2         b >/ 2         a >/3

for every 2-power team  (a b c d).

===============================================

A sharp inequality
--------------------------

Inequality  (i)  is a part of the definition.  Inequality (i')  is  slightly sharper. Now we will prove one that is still clearly sharper.

Let  (a b c d)  be a 2-power team. Then

                        c^2 - d^2  =  a^2 - b^2 = (a-b)*(a+b)
hence
                       --------------------------------------------------------
                       2*(a-b)*b   <   c^2 - d^2  <  2*(a-b)*a
                       --------------------------------------------------------

===========================================

c-d  >/  3
-----------

Let  (a b c d)  be an arbitrary 2-power team. We know that  c-d >/ 2, and  1 \< a-b < c-d.  Thus,

         c-d = 2  ==>  a =1


But  c-d  and  a-b  must be of the same parity -- a contradiction that proves:

Theorem 1   Difference  c-d >/ 3  for every 2-power team  (a b c d).

Thus, inequality (i')  admits an improvement:

(i'')     a  >  b  >/  c  >/  d+3  >/  3               for every team  (a b c d)

Thus,
                  d >/ 0         c >/ 3         b >/ 3         a >/4

for every 2-power team  (a b c d). 

Now, of course:

    a = 4   ==>   b=3  c=3  d=0

However  then

      a^2+d^2 = 4^2 + 0^2  =  16  <  3^2+3^2 = b^2+c^2

hence   a^2+d^2 < b^2 + c^2  in this case. Thus  (4 3 3 0)  is not a team. It follows that

Theorem 2   If  (a b c d)  is a 2-power team then  a >/ 5.

===========================================

Minimal leader  a=5
--------------------------

Let  A := (a b c d)  be an arbitrary 2-power team.  If  a =5  then by team inequality  (i'')
there are only two quadruples A = (5 b c d) that have a chance to be teams, namely:

                a=5   >   b=3 or 4   >/   c=3   >   d=0

However  5-3  had different parity from  3-0  hence quadruple  (5 3 3 0)  is not a team.

The last candidate team  (5 4 3 0)  actually is a team since

              5^2 + 0^2  =  4^2 + 3^2

Theorem 3   Among all 2-power teams  (a b c d)  the smallest leader is  a=5. Furthermore,
a = 5 is a leader of one team only, namely of  (5 4 3 0).

===========================================

a=6  is not a leader
------------------------

Consider an arbitrary integer quadruple  (a b c d)  such that  (i")is satisfied, and  a=6.  Then

    b=5  and c >/ 4   ==>  c^2-d^2 = (c-d)*(c+d) >/ 3*4 > 11 = a^2-b^2

         ==>  c^2-d^2  >  a^2-b^2  ==>  a  is not a leader (under the circumstances)

Next,
            b \< 4  ==>  a^2 - b^2 >/ 20 > 4^2 - 0^2 >/  c^2 - d^2

         ==>  c^2-d^2  <  a^2-b^2  ==>  a  is not a leader (under the circumstances)

Thus  a = 6  is never a 2-power team leader.

===========================================

Some non-team cases
---------------------------

Theorem 4   If  a+b  or  a+c  is prime then the integer quadruple  (a b c d)  is not a 2-power team;

Proof   A proof by contradiction, the prime  a+c  case:

         a + c | a^2 - c^2 = b^2 - d^2 = (b-d)*(b+d)

If  a+c. were prime then  a+c | b-d  or. a+c | b+d.  However

              a + c > b + d >/ b - d

-- a contradiction.

The prime  a+b  case is similar.  End of PROOF

---

Theorem 5   If  a+b  or  a+c  is squarefree then the integer triple  (a b c)  is not a Pythagorean triangle, i.e  a^2 =/= b^ + c^2 (in other words  (a b c 0) is not a 2-power team).

Proof   In general, if integer  x  is square free, and  x | c^2  then  x | c.

Thus, the  a+b  case proved by a contradiction:

            a^2 = b^2 + c^2   ==>   a+b | a^2-b^2 = c^2

If  a+b  were squarefree then then  a+b | c -- the case is proven.

The squarefree  a+c  case is similar.  End of PROOF

===========================================

b=c case, and sqrt(2)
---------------------------

Let  (a b c d)  be a 2-power team such that  b=c so that

                        a^2 + d^2  =  2*c^2
i.e.
               (a/c)^2  = 2 - (d/c)^2  =  (sqrt(2) - d/c)*(sqrt(2) + d/c)
                                =  (sqrt(2) - d/c)^2 * (sqrt(2) + d/c) / (sqrt(2) - d/c)
hence
                         sqrt(2)- d/c  <  a/c  <   sqrt(2)

Especially, when  d=1  the above approximations  a/c  of  sqrt(2)  are extra sharp (for large  a c), and there occurs infinitely many pairs  (a c)  that solve equation

                                 a^2 + 1  =  2*c^2

The smallest solution of this equation, namely  (a c) := (1 1)  (so that 1^2+1=2*1^2), doesn't count since then
                         (a b c d) := (1 1 1 1)
is not a team. The smallest one that leads to a 2-power team is  (a c) := (7 5),  then

                         (a b c d) := (7 5 5 1)
is a team, and
                                       7^2 + 1 = 2*5^2        so that    7^2 + 1^2  = 5^2 + 5^2

Thus

Theorem 6    a=7  is the smallest leader among all teams such that  d > 0.

Given an  (a b c d)  team as above we obtain the next  (a' b' c' d')  team by recursion:

                a' := 3*a +4*c        &        c' :=  2*a + 3*c

That's how team  (7 5 5 1)  can be obtained from  (1 1 1 1), and we can obtain the next from (7 5 5 1), namely  (41 29  29  1):

                  41^2 + 1  =  2*29^2                     so        41^2 + 1^2  =  29^2 + 29^2
etc.

Remark 1   Equation   a^2 - 1 = 2*c^2   is about equally attractive as  a^2 + 1  =  2*c^2  that we discussed
 above. The total of the solutions of both equations is catched exactly by a single equation

                              | a^2 - 2*c^2 | = 1

The  a/c  fractions provide the best rational approximations for  sqrt(2).  The combined set of. (a c)-solutions is obtained from  (a c) := (1 1)  by iterating the equations:

                a' := a +2*c        &        c' :=  a + c

This kind of equations and iterations were studied first by the Indian mathematicians in the early medieval times, then by Europeans during the late medieval times.

===========================================

A 3-parameter family of 2-power teams
---------------------------------------------------

Consider arbitrary three natural numbers  n  s  t  subjected only to the following constrains:

                     n > t        &        s > 1
Then
            ((1/2)*(n*s + t))^2  +  ((1/2)*(n-s*t))^2   =   ((1/2)*(n + s*t))^2  +  ((1/2)*(n*s - t))^2

Let's establish the additional conditions that make the absolute values of the above squared expressions (before they are squared) into a team. The first one demand is that they are integers. Thus let

(iii)   either all three integers   n s t  are even   or   s  is odd and  n t  are of the same parity

Next observe that

  • n*s + t  >  n+s*t

  • n*s + t  > n*s - t  >  0

  • n + s*t  >  | n-s*t |

  • n*s - t  >  | n-s*t |
About the last inequality:

       | n-s*t | = n-s*t   ==>   (n*s-t) - |n-s*t|  =  n*(s-1) - t*(s-1) = (n-t)*(s-1) > 0
and
       | n-s*t | = s*t-n   ==>   (n*s-t) - |n-s*t|  =  n*(s+1) - t*(s+1) = (n-t)*(s+1) > 0

Great!

Thus, define:

(iv)   a := (n*s+t)/2       b := max(n+s*t   n*s-t)      c := min(n+s*t   n*s-t)      d := | n - s*t |

Now, after (iv),

                         (a b c d)  is a 2-power team

===========================================

Every  a >/ 7  is a leader
-------------------------------

Assume (iv).

Let's start with some small warm-up examples that present team equation  a^2 + d^2 = b^2 + c^2:

Examples 0
  • n=s=3  t=1  ==>  5^2 + 0^2  =  4^2  + 3^2

  • n=4  s=t=2  ==>  5^2 + 0^2  =  4^2  + 3^2

  • n=4  s=3  t=2  ==>  7^2 + 2^2  =  5^2 + 5^2

  • n=5  s=3  t=1  ==>  8^2 + 1^2  =  7^2 + 4^2

  • n=3  s=5  t=1  ==>  8^2 + 1^2  =  7^2 + 4^2

  • n=s=4  t=2  ==>  9^2 + 2^2  =  7^2 + 6^2

Notation   Integer  r(x) = 0 or 1, and  2 | x - r(x).

Theorem 7  Every integer  a >/ 7  is leading a proper 2-power team  (a b c d).

Proof       Thus, let  integer a >/ 7,   and
  •      s := 2
and define  n t  in terms of  a:
  •      t  :=  2*(2-r(a))

  •      n  :=  a - t/2  =  a + r(a) - 2
Then

            (n*s+t)/2  =  n + t/2  =  a

 in an agreement with (iv).  Also, all three variables  n s t  are even, where  s>1.  And, finally,

                                  t \< 4
while
                   n  =  a - t/2  >/  7 - 2  =  5  >  t

Thus we can define  b c d  by (iv), it produces proper 2-power team (a b c d) for every integer  a >/ 7. All these teams are proper since

            d := | n - s*t | = | (a - t/2) - 2*t | = | a - 5*t/2 | >/ a - 5*t/2 >/ 7 - 4 = 3 > 0

End of PROOF

Now, based on the earlier discussion, let me formulate a complete statement:

Theorem 7'  The set of all leaders  a  of 2-power teams consists exactly of  a=5,

and of all  a >/ 7.

Moreover, the set of leaders of all proper teams consists exactlu of all integers a >/ 7.



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