Define LT(x)=last two digits of x (so LT(100)=00, LT(234)=34, ect.). Then LT(2^5+3^5)=75. Define $\displaystyle {k}^{*}$ to be the smallest integer>5 such that 1) LT(2^$\displaystyle {k}^{*} $+3^$\displaystyle {k}^{*} $)=00 or 25 or 50 or 75; 2) LT(2^$\displaystyle {k}^{*} $)=LT(2^5); 3) LT(3^$\displaystyle {k}^{*} $)=LT(3^5). Such a $\displaystyle {k}^{*} $ is found by inspecting the first 20 terms of the table to be $\displaystyle {k}^{*} =25$. And LT(2^$\displaystyle {k}^{*} $+3^$\displaystyle {k}^{*} $)=75. No problem with this?

Now because LT function takes only the last two digits, it is true that LT(2^(5+i))=LT$\displaystyle ({2}^{5}{2}^{i})$=LT(LT(2^5)LT(2^i))=LT(LT(2^$\displaystyle {k}^{*}$)LT(2^i))=LT($\displaystyle {2}^{{k}^{*}}{2}^{i}$)=LT($\displaystyle 2^{{k}^{*}+i}$), for any i>0.

Similarly LT(3^(5+i))=LT($\displaystyle 3^{{k}^{*}+i}$), too. So LT(2^(5+i)+3^(5+i))=LT(LT(2^(5+i))+LT(3^(5+i)))=LT (LT(2^($\displaystyle {k}^{*}+i$))+LT(3^($\displaystyle {k}^{*}+i$)))=LT(2^($\displaystyle {k}^{*}$+i)+3^($\displaystyle {k}^{*}$+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($\displaystyle {k}^{*}-1$)}, and any f(k)$\displaystyle \in${00,25,50,75} iff there exists 5$\displaystyle \leq$k'$\displaystyle \leq$$\displaystyle {k}^{*}-1$ such that f(k')=f(k).

So what is this k' if f(k')$\displaystyle \in${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.