Equation with Two Exponential Terms
Date: 06/27/2009 at 15:34:36 From: Dan Subject: find ordered pair (a,b) for 3^a + 7^b giving a perfect square Any perfect square can be expressed as n^2, find all ordered pairs for 3^a + 7^b which result in a perfect square answer. I'm not sure where to begin. I have an idea that there are several situations to analyze. For example if a and be are equal they can be factored by a formula and if they are both even they are going to be complex roots.
Date: 06/29/2009 at 08:36:01 From: Doctor Jacques Subject: Re: find ordered pair (a,b) for 3^a + 7^b giving a perfect square Hi Dan, This is a very nice exercise about congruences. Although the proof does not require any elaborate concept, it requires a significant amount of work--be prepared for a hard ride... We are looking for solutions of: 3^a + 7^b = n^2  where a, b, and n are non-negative integers. We can first look at  as a congruence modulo 7: 3^a = n^2 (mod 7)  Modulo 7, the quadratic residues (i.e., the squares) are 0, 1, 2, and 4, and 3 is not one of them. And we also have 3^6 = 1 (mod 7)  by Fermat's theorem. This implies that a must be even. [We could also look at congruences modulo 4 to deduce that b must be odd, but we will not need that fact.] As a is even, we may write a = 2k for some integer k. We have: 3^(2k) + 7^b = n^2 n^2 - 3^(2k) = 7^b (n + 3^k)(n - 3^k) = 7^b  This means that the product on the LHS must be a power of 7, and therefore each factor must also be a power of 7. However, it is not possible to have both factors on the LHS divisible by 7, as this would imply that their difference (n + 3^k) - (n - 3^k) = 2*3^k would be divisible by 7, which is clearly impossible. The only power of 7 that is not a multiple of 7 is 7^0 = 1. We may therefore conclude that one of the factors must be 1, and this is obviously the smaller factor; we have: n - 3^k = 1 n = 3^k + 1  If we substitute that into , we obtain: 2*3^k + 1 = 7^b 2*3^k = 7^b - 1  and this means that 7^b - 1 must be twice a power of 3. Let us look at the simple cases first. If k = 0, we have 7^b - 1 = 2, which is impossible. If k = 1, we have: 6 = 7^b - 1 which has the solution b = 1. We will now show that this is the only solution. We may now assume that k > 1, and  shows that 7^b - 1 must be divisible by 9; in other words, we must have: 7^b = 1 (mod 9)  Modulo 9, the powers of 7 are: m | 0 1 2 3 ----+-------- 7^m | 1 7 4 1 and this shows that  will be satisfied if and only if b is a multiple of 3. We have: 7^3 - 1 = 342 = 2*3^2*19 but this means that: 7^3 = 1 (mod 19) and we will have 7^b = 1 (mod 19) for all exponents b that are multiples of 3; this is the same as saying that, if b is a multiple of 3, 7^b - 1 is a multiple of 19, which contradicts equation . Therefore, the only solution is k = 1, a = 2 and b = 1, corresponding to 3^2 + 7^1 = 4^2. Please feel free to write back if you require further assistance. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/
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