Diophantine Equation 1

[MM]

The following problem was proposed on the Bulgarian Spring Mathematical Competition 2007 for 8th grade.
Solve in natural numbers [tex]3^{m}=26^{n}+53[/tex].

[broniran]

The last digit of [tex]26^n[/tex] always is [tex]6[/tex], so the last digit of [tex]3^m[/tex] must be 9. Since the last digit of [tex]3^m[/tex] is 9, [tex]m[/tex] is even. The left side is [tex]\equiv 0(mod 3)[/tex], so the right should also be [tex]\equiv 0(mod3)[/tex]. We that [tex]53\equiv-1(mod3)[/tex] so [tex]26^n[/tex] must be [tex]\equiv 1(mod3)[/tex] , which leads to the conclusion that [tex]n[/tex] is even, too. Let [tex]m=2t[/tex] and [tex]n=2k[/tex]. Now the equation is equivalent to:
[tex](3^t-26^k)(3^t+26^k)=53[/tex]
But since [tex]53[/tex] is prime and [tex]3^t -26^k<3^t+26^k[/tex], we have:
[tex]3^t-26^k=1, 3^t+26^k=53[/tex]
And we easily obtain[tex]t=3,k=1\Rightarrow m=6,n=2[/tex]
[tex]3^6=26^2+53[/tex] is the only answer

[MM]

Any other solutions?

[shahjee1]

Rock'n'roller - The approach was great, and the result was correct ... but you either started with the wrong formula, or applied it wrongly ... then had to fudge the intermediate result (+9 magically got to the numerator of a fraction with the denominator 9, also as 9, instead of 81) ... just to make the result correct.