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Unequally Yoked III (Posted on 2009-09-01) Difficulty: 3 of 5
Determine all possible nonzero integer(s) P and Q, with PQ, that satisfy this equation.

                                  PP+2Q = QQ+2P

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Some Thoughts computer exploration (possible spoilers) | Comment 1 of 3

Since the equation is symmetrical with regard to p and q, we can assume abs(p)<=abs(q), and interchange the p and q of any results.

list
   10   for T=1 to 6000
   20    for P=1 to int((T)/2)
   30       Q=T-P:Np=-P:Nq=-Q
   40       if P<Q then if P^(P+2*Q)=Q^(Q+2*P) then print P;Q
   41       if Np^(Np+2*Q)=Q^(Q+2*Np) then print Np;Q
   42       if P<Q then if Np^(Np+2*Nq)=Nq^(Nq+2*Np) then print Np;Nq
   43       if P^(P+2*Nq)=Nq^(Nq+2*P) then print P;Nq
   50    next
   60   next
OK
run
 1 -2
-4  32
 4 -32
 16  32
-16 -32
 16  64
-16 -64
Overflow in 42
?t
 664
OK

This has ostensibly found all the solutions where the total of the absolute values of p and q is under 664. Keep in mind that the p and q values can be interchanged.

I use the word "ostensibly", as rounding errors for when the exponents are negative may produce mismatches when in theoretical fact the two values are equal (lhs and rhs).


  Posted by Charlie on 2009-09-01 12:35:34
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