| Let N denote the set of positive integers.It is well known that any positive integer n can be uniquely expressed n-a0+a1b+…am+bm,where integer b>1 is the integer basis,ai is the coefficient,a,∈{0,1,…,b-1},0≤i≤m.Given an integer basis b>1,Let Bk(b)represent the set of positive integers in the above equation where the number of non-zero integers in the coefficients al,a2,…,am happen to be k.For a particular sequence of positive integers S,in general,the intersection S ∩Bk(b)is a finite set.But proving this conclusion is very difficult.If S consists of all squares,then S∩Bk(b)is not finite((1+bl)2=1+2bl+b2l,l∈N).In 2013,M.Bennett studied the case of b=3 and proved some conclusions.When classifying the ternary expansion of n5,Bennett studied equations 2δ3a+2δ2=n5,a>0,δi∈{0,1}and 2δ13a+2δ23b+2δ3=n5,a>b>0,δi∈{0,1}.He proved that the equations have no solution except(δ1,δ2,δ3)=(0,0,1).Therefore,only the case of(δ1,δ2,δ3)=(0,0,1)remains,that is 3a+3b+2=n5.In the same year,S.Singh studied equation 3a+3b+2-n5 and obtained the following conclusion:If a≥b>0,n∈N,then Diophantine equation 3a+3b+2=n5 has no positive integer solution when 2<n≤2+6·106.Based on this,we will use the elementary methods of factorization,congruence and the property of 3-adic and diophantine approximations,proving that(a,b,n)=(3,1,2)is the only positive integer solution of diophantine equation 3a+3b+2-n5 with a≥b>0,n∈N. |
