Well, LTE lemma says (because 1991 is odd) that $v_5(102^{1991}+103^{1991})=v_5(102+103)+v_5(1991)=v_5(205)+0=1$, so 5 divides our number, but no power of 5 does, so our number can't be a proper power. Q.E.D.

hi, what does LTE stands for? L______ T______ E_______ ohh! lifting the exponent! (seriously, i just figured it out after writing those blanks out )

Lifting the exponent lemma (i think)

forthegreatergood wrote: Lifting the exponent lemma (i think) yes, see my edit. thanks anyways!

Note that by taking modulo $5$, \[102^{1991}+103^{1991}\equiv 2^{1991}+(-2)^{1991}\equiv 0 \ \textrm{(mod $5$)}\]and taking modulo $41$, \[102^{1991}+103^{1991}\equiv (-21)^{1991}+21^{1991}\equiv 0 \ \textrm{(mod $41$)}.\]However, taking modulo $25$, we have \[102^{1991}+103^{1991}\equiv \frac{1}{2}+\frac{1}{3}\equiv 20 \ \textrm{(mod $25$)}.\]Thus, it can't be a proper power of an integer. $\blacksquare$

Sorry, but what even means “proper power of an integer”.

rafaello wrote: Sorry, but what even means “proper power of an integer”. one where the exponent is integer ( aka a perfect power) ?