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How would I go about finding the solutions to the exponential diophantine equation $18n+10=2^k$ ?

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Hint: You want $2^k \equiv 10 \mod 18$. Certainly $2^k \equiv 0 \equiv 10 \mod 2$ for $k \ge 1$, so it is sufficient for $2^k \equiv 1 \equiv 10 \equiv \mod 9$. So try a few powers of $2$ until you find one congruent to $1$ mod $9$...

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  • $\begingroup$ I've done that and I've ended up with all k which are multiples of 6 being solutions - any idea as to why this is? $\endgroup$ – John Jul 23 '12 at 5:33
  • $\begingroup$ @Matt : Because the elements coprime to $m$ in ${\mathbb Z}_m$ form a group under multiplication. You might look up en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n $\endgroup$ – Robert Israel Jul 23 '12 at 6:39
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This problem can be simplified to --> 9n+5=2^k1 (9m+1)*(9n+5)=81*m*n + (9n+45m)+5 =9m'+5 ,where m'=9*m*n+n+5m If 9m+1=2^k2 , then 9m'+5 =2^(k1+k2) for n=3 k1=5 and for m=7 k2=6 are solutions so 9m'+5 =2^11 is a solution 2^11-5 =2043 =9*227 --This process can be repeated.

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  • $\begingroup$ Please consider editing your answer and use $\LaTeX$. For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – Stefan Hansen Feb 14 '13 at 9:29
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Given,

$18n + 10 = 2^k$

i.e. $$[(2)(3^2)(n) + (2)(5)] = 2^k$$

i.e. $$(2)[(3^2)(n) + (3+2)] = 2^k$$

i.e. $$(2)[(3)(3)(n) + (3) + 2] = 2^k$$

i.e. $$(2)[(3)((3)(n) + 1) + 2] = 2^k$$

i.e. $2^{3 + 1 + 2}$ is a solution of this equation for $k = (3 + 1 + 2)$

i.e. $18n + 10 = 2^{3 + 1 + 2}$

i.e. $18n + 10 = 64$

i.e. $18n = 54$

i.e. $k=6, n=3$ is a solution of the equation $18n + 10 = 2^k$

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