Guest Posted August 1, 2009 Report Share Posted August 1, 2009 Each of M and N is a positive integer. Determine all possible value(s) of a positive prime constant P such that the equation: 2M – 3*P = N2 has precisely two distinct solutions in (M, N). Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted August 1, 2009 Report Share Posted August 1, 2009 (edited) Each of M and N is a positive integer. Determine all possible value(s) of a positive prime constant P such that the equation: 2M – 3*P = N2 has precisely two distinct solutions in (M, N). There are none. N can't be 0 mod 3 as that would imply that 2^M is divisible by 3. If N is 1 mod 3, then 3 must divide 2^M-1 and so M must be even (it is easy to show that 3 divides 2^X-1 iff X is even). If N is 2 mod 3, then 3 must divide 2^(M-2)-1 and again M must be even. So, M is even. Then, we can rearrange and factor to get: (2^(M/2)-N)*(2^(M/2)+N)=3*P So, either [(2^(M/2)-N)=P and (2^(M/2)+N)=3] or [(2^(M/2)-N)=3 and (2^(M/2)+N)=P]. In either case, setting P determines both N and M uniquely. Therefore, there is no P which will give two distinct solutions in (N,M). Edited August 1, 2009 by superprismatic Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 2, 2009 Report Share Posted August 2, 2009 (edited) There are none. N can't be 0 mod 3 as that would imply that 2^M is divisible by 3. If N is 1 mod 3, then 3 must divide 2^M-1 and so M must be even (it is easy to show that 3 divides 2^X-1 iff X is even). If N is 2 mod 3, then 3 must divide 2^(M-2)-1 and again M must be even. So, M is even. Then, we can rearrange and factor to get: (2^(M/2)-N)*(2^(M/2)+N)=3*P So, either [(2^(M/2)-N)=P and (2^(M/2)+N)=3] or [(2^(M/2)-N)=3 and (2^(M/2)+N)=P]. In either case, setting P determines both N and M uniquely. Therefore, there is no P which will give two distinct solutions in (N,M). I cannot refute the methodology corresponding to the foregoing proof. However, P=5 seems to satisfy all the conditions of the given problem due to the following reasons: The equation becomes 2^M – 15 = N^2 at P = 5. Reducing this to mod 3, it can easily be shown that: M must be even, so that: (2^M/2 + N) (2^M/2 - N) = 15. Since, M and N are positive integers it follows that: 2^M/2 + N > 2^M/2 – N, and accordingly: (2^M/2 + N, 2^M/2 - N) = (15, 1), (5, 3), giving: (2^M/2, N) = (8, 7), (4, 1), so that: (M, N) = (6, 7), (4, 1). Consequently, it seems that there exists at least one value of P that satisfies all the given conditions. Although I came across this problem in an old mathematics periodical, the said magazine does not have the answer to this problem. Till date, I have not yet been able to come up with a proof giving all the value(s) of P and I'm still looking for the same. Edited August 2, 2009 by K Sengupta Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted August 2, 2009 Report Share Posted August 2, 2009 I cannot refute the methodology corresponding to the foregoing proof. However, P=5 seems to satisfy all the conditions of the given problem due to the following reasons: The equation becomes 2^M – 15 = N^2 at P = 5. Reducing this to mod 3, it can easily be shown that: M must be even, so that: (2^M/2 + N) (2^M/2 - N) = 15. Since, M and N are positive integers it follows that: 2^M/2 + N > 2^M/2 – N, and accordingly: (2^M/2 + N, 2^M/2 - N) = (15, 1), (5, 3), giving: (2^M/2, N) = (8, 7), (4, 1), so that: (M, N) = (6, 7), (4, 1). Consequently, it seems that there exists at least one value of P that satisfies all the given conditions. Although I came across this problem in an old mathematics periodical, the said magazine does not have the answer to this problem. Till date, I have not yet been able to come up with a proof giving all the value(s) of P and I'm still looking for the same. Thanks, you helped me find the kink in my reasoning! Thanks, you have found the flaw in my argument! I didn't consider one other case, namely, that (2^(M/2)-N)=1 and (2^(M/2)+N)=3*P. So, when P=5, you get the (8,7) answer and from the (2^(M/2)-N)=3 and (2^(M/2)+N)=P possibility, you get the (4,1) answer. So, that means that this pair of possibilities are the only way you can get 2 answers (the other case, when [(2^(M/2)-N)=P and (2^(M/2)+N)=3], is actually inconsistent because P must be 2 -- the only prime less than 3). Anyway, in my overlooked case, (2^(M/2)-N)=1 and (2^(M/2)+N)=3*P, P must be of the form 1+2^2+2^4+2^6+...+2^(2n) for some n. 5 is the case when n=1. If there are any other primes of this form, then there are other answers with 2 distinct solutions in (M,N). I checked up to n=20 with no luck. If I find another, I'll be sure to let you know. Thanks for the wonderful problem and for giving me the hint to the flaw in my "proof"! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 2, 2009 Report Share Posted August 2, 2009 (edited) And it is now possible to determine that P can only be 5, which solves Sengupta's original puzzle. We have already found from previous posts that, if 2^M - 3P = N^2, then M must be even so that, putting M = 2m (2^m - N)(2^m + N) = 3P, an equation which can only be solved if (a) 2^j - N = 3 and 2^j + N = P so that 2^(j+1) = P + 3 and/or (b) 2^k - N = 1 and 2^k + N = 3P so that 2^(k+1) = 3P + 1. If the problem as stated possesses two solutions for the same prime number P, then 3 * 2^(j+1) - 2^(k+1) = 8 i.e. 3 * 2^(j-2) - 2^(k-2) = 1. Clearly the only solution to this Diophantine equation is j = 2 and k = 3. This immediately yields P = 5. Edited August 2, 2009 by jerbil Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted August 2, 2009 Report Share Posted August 2, 2009 And it is now possible to determine that P can only be 5, which solves Sengupta's original puzzle. We have already found from previous posts that, if 2^M - 3P = N^2, then M must be even so that, putting M = 2m (2^m - N)(2^m + N) = 3P, an equation which can only be solved if (a) 2^j - N = 3 and 2^j + N = P so that 2^(j+1) = P + 3 and/or (b) 2^k - N = 1 and 2^k + N = 3P so that 2^(k+1) = 3P + 1. If the problem as stated possesses two solutions for the same prime number P, then 3 * 2^(j+1) - 2^(k+1) = 8 i.e. 3 * 2^(j-2) - 2^(k-2) = 1. Clearly the only solution to this Diophantine equation is j = 2 and k = 3. This immediately yields P = 5. Nice finish, Jerbil! Your completion also means (see my previous observations) that the only prime of the form 1+sum{i=1 to n}(4^i) is 5. Now, we can have a real number theory type proof of this fact starting with "For a given prime P, consider multiple solutions of 2^M - 3P = N^2 ....". Quote Link to comment Share on other sites More sharing options...
Question
Guest
Each of M and N is a positive integer. Determine all possible value(s) of a positive prime constant P such that the equation: 2M – 3*P = N2 has precisely two distinct solutions in (M, N).
Link to comment
Share on other sites
5 answers to this question
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.