All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Unequally Yoked III (Posted on 2009-09-01)
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)

Comments: ( Back to comment list | You must be logged in to post comments.)
 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

 Search: Search body:
Forums (0)