Define LT(x)=last two digits of x (so LT(100)=00, LT(234)=34, ect.). Then LT(2^5+3^5)=75. Define

to be the smallest integer>5 such that 1) LT(2^

+3^

)=00 or 25 or 50 or 75; 2) LT(2^

)=LT(2^5); 3) LT(3^

)=LT(3^5). Such a

is found by inspecting the first 20 terms of the table to be

. And LT(2^

+3^

)=75. No problem with this?

Now because LT function takes only the last two digits, it is true that LT(2^(5+i))=LT

=LT(LT(2^5)LT(2^i))=LT(LT(2^

)LT(2^i))=LT(

)=LT(

), for any i>0.

Similarly LT(3^(5+i))=LT( ), too. So LT(2^(5+i)+3^(5+i))=LT(LT(2^(5+i))+LT(3^(5+i)))=LT (LT(2^( ))+LT(3^( )))=LT(2^( +i)+3^( +i))
From this we can see f(k)=LT(2^k+3^k) takes only FINITELY MANY values for k=5,6,7,...

and the values it can take are exactly equal to this set:{f(5),f(6),f(7),...,f(

)}, and any f(k)

{00,25,50,75} iff there exists 5

k'

such that f(k')=f(k).

So what is this k' if f(k')

{00,25,50,75}? Inspect the table we find k'=5 or k'=15.

So now you see where k=5+20m and k=15+20m come from.

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If this doesn't convince you. I happily quit

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EDIT: red letters.