P is prime 2 ^ P + 3 ^ P = a ^ n prove n = 1

P is prime 2 ^ P + 3 ^ P = a ^ n prove n = 1

Let (K + 6) ^ 2 = n + 36 = 2 ^ A * 3 ^ B. then K (K + 12) = 2 ^ A * 3 ^ B
Let k = 2 ^ c * 3 ^ D, then 2 ^ c * 3 ^ D + 2 ^ 2 * 3 = 2 ^ e * 3 ^ F
When d = 0, if C & lt; 2, then there is no solution
If C & gt; = 2,
Let P = C-2, s = E-2, t = F
There are 2 ^ P + 3 = 2 ^ s * 3 ^ T. (III)
When S & gt; = P, then p = 0, s = 2, t = 0, n = 2 ^ 6 = 64
When S & lt; P, then s = 0,2 ^ P + 3 = 3 ^ t
When D & gt; = 1, if C & lt; 2,
Let P = 2-C & gt; 0, q = D-1 & gt; = 0, s = E-C, t = F-1, have 3 ^ Q + 2 ^ P = 2 ^ s * 3 ^ t
If P & lt; = s, then p = 0, (rounding off)
If P & gt; s, then s = 0,3 ^ Q + 2 ^ P = 3 ^ t
If Q & lt; = t, then q = 0,1 + 2 ^ P = 3 ^ t, because 0 & lt; P & lt; = 2, so p = 1, t = 1,
c=1,d=1,k=2*3=6,n=2^2*3^3=108.
If D & gt; = 1 and C & gt; = 2
Let P = C-2, q = D-1, s = E-2, t = F-1
There are 2 ^ p * 3 ^ Q + 1 = 2 ^ s * 3 ^ t
If S & gt; = P, then 1 = 2 ^ p * (2 ^ (S-P) * 3 ^ T-3 ^ q),
1=2^(s-p)*3^t-3^q.
If T & gt; = q, then q = 0,2 ^ s * 3 ^ t = 2, s = 1, t = 0
k=2^2*3=12,n=2^5*3^2=288.
If T & lt; Q, then t = 0,2 ^ S-3 ^ q = 1, Q & gt; 0
If S & lt; P, then s = 0
If T & gt; = q, then q = 0, - 2 ^ P + 3 ^ t = 1. T & gt; = 0
If T & lt; Q, then t = 0,2 ^ p * 3 ^ Q + 1 = 0, there is no solution
Therefore, the original problem is transformed into a solution
2 ^ s = 3 ^ Q + 1 and 2 ^ P + 1 = 3 ^ t
For 2 ^ s = 3 ^ Q + 1, Q & gt; 0 (II)
S = 2, q = 1 is a solution,
So n = 2 ^ 6 * 3 ^ 3 = 1728
For 2 ^ P + 1 = 3 ^ t, T & gt; = 0 (I), P = 1, t = 1 is the solution N = 2 ^ 5 * 3 ^ 3 = 864
P = 3, t = 2 is the solution, n = 2 ^ 7 * 3 ^ 4 = 10368
It can be proved that when T & gt; 2, (I) has no solution; when Q & gt; 1, (II) has no solution
So n can be 64108288172886410368
It is proved that when T & gt; 2, 2 ^ P + 1 = 3 ^ t has no solution; when Q & gt; 1, 2 ^ s = 3 ^ Q + 1 has no solution
Because T & gt; 2, then 9|2 ^ P + 1. And 2 ^ P + 1 = (3-1) ^ P + 1 = 9W + p * (- 1) ^ (p-1) * 3 + (- 1) ^ P + 1 = 9W + V
If P is odd, then v = 3P-1 + 1 = 3P, the solution is indeterminate
If P is even, then v = - 3P + 1 + 1,9! | V, so there is no solution
Because when Q & gt; 1, S & gt; 2,2 ^ S-1 = (3-1) ^ S-1 = 9W + s * (- 1) ^ (s-1) * 3 + (- 1) ^ S-1 = 9W + v.9|2 ^ s + 1
If s is odd, then v = 3p-1-1 = 3p-2,9! | V, so there is no solution
If s is even, then v = - 3P + 1 - 1 = 3P, the solution is indeterminate
2^p+1=3^t(I),2^s-1=3^q(II)
If Q & lt; = t, then S & lt; = P, then (2 ^ s-1) | (2 ^ P + 1)
The remainder of 2 ^ P + 1 divided by 2 ^ S-1 is w = P% s, W & lt; s, 2 ^ W + 1 = 2 ^ s-1
We get w = 1, s = 2
It is verified by computer that (I) and (II) have no other solutions within 100