Problems & Puzzles: Conjectures Conjecture 40. Karama's Conjecture Regarding the above equation, Muneer Jebreel Karama conjectures that: 1) there is no solution if p, q, r, s & t are prime numbers. On the other hand, he has shown that: 2) if the prime condition is eliminated the same equation has infinite solutions. Question: Can you verify both claims from Karama?
Contributions came from Patrick Capelle, Rudolph Knjzek, T.D. Noe, Dan Dima & Adam Stinchcombe. *** Capelle wrote: The first conjecture is false. If you take p = q = r = 2 and s = t = 3, you obtain: 3^{2} + 3^{2} + 3^{2} + 3^{3} + 3^{3} = 81 = 9^{2} *** Knjzek wrote:
*** T. D. Noe wrote:
*** Dan Dima wrote: 3^{p} + 3^{q} + 3^{r} + 3^{s} + 3^{t} = k^{2}. 1) If p, q, r, s & t are all prime numbers the only solution is: (3,3,2,2,2) & k=9 since 3^{3} + 3^{3} + 3^{2} + 3^{2} + 3^{2} = 9^{2} and the permutations. Only if p, q, r, s & t are distinct prime numbers the conjecture is true, but generally this is not true so the conjecture is false! Proof. Suppose p ³ q ³ r ³ s ³ t ³ 2 all primes. Note that k must be odd and for any odd k^{2} º 1 (mod 8) and 3^{k} º 3 (mod 8).
p2 is odd so 3^{p2} º 3 (mod 8) Þ m^{2} º 7 (mod 8), contradiction Þ no solution. 3. p ³ q > r = s = t = 2 Þ 3^{p2} + 3^{q2} + 3 = m^{2}. p2, q2 both odd so 3^{p2} º 3^{q2} º 3 (mod 8) and the equality may hold mod 8: 3^{p3} + 3^{q3} + 1 = 3n^{2} Þ p = q = 3, otherwise by contradiction it cannot hold mod 3. Hence (3,3,2,2,2) is such a solution. 4. p ³ q ³ r > s = t = 2 Þ 3^{p2} + 3^{q2} + 3^{r2} + 2 = m^{2}. p2, q2, r2 odd so m^{2} º 3 (mod 8), contradiction Þ no solution.
m^{2} º 5 (mod 8), contradiction Þ no solution.
m^{2} º 7 (mod 8), contradiction Þ no solution. Hence if p, q, r, s & t are all prime numbers there is a solution (3,3,2,2,2). More generally this is the only solution if we consider p, q, r, s & t odd numbers or 2. 2) If the prime condition is eliminated one can easily find many different parameterizations to show that it has infinite solutions. If (p, q, r, s, t, k) is a solution then (p+2, q+2, r+2, s+2, t+2, 3k) is also a solution. I think it is quite impossible to determine all the solutions for this equation. Some parameterizations are:
(2i+1, 2i+1, 2i, 2i, 2i, 3^{i+1}) ***





