[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: 2^M – 3*P = N² has precisely two distinct solutions in (M, N).

[superprismatic]

Each of M and N is a positive integer. Determine all possible value(s) of a positive prime constant P such that the equation: 2^M – 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

[Guest]

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

[superprismatic]

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"!

[Guest]

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

[superprismatic]

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 ....".