\[Let \quad a^n+b^n=c^n \quad be \quad true \quad for\quad \forall n\geqslant 3\] \[\textit{then there exists a positive integer k such that}\quad c<c^2<c^3<....<c^k<a^n <b^n\] \[and \quad c<c^2<c^3<....<c^k<a^n <b^n<c^{k+1}\] \[If \quad a^n+b^n=c^n \quad is \quad true \quad \forall n\geqslant 3 \quad then\] \[a^n=cq_1+r_1\quad \textit{by Euclid's Division Lemma}\quad 0\leqslant r_1<c\] \[b^n=cq_2+r_2\quad \textit{by Euclid's Division Lemma}\quad 0\leqslant r_2<c\] \[a^n+b^n=cq_1+r_1+cq_2+r_2\] \[\quad 0\leqslant r_1+r_2<2c\] \[\therefore c^n=a^n+b^n=c(q_1+q_2)+r_1+r_2\quad \because a^n+b^n=c^n \] \[c^{n}=cq_1+r_1+cq_2+r_2\quad \] \[\therefore c^n=c\left\{q_1+q_2+\left(\frac{r_1+r_2}{c}\right)\right\}\quad \] \[\therefore c^{n-1}=\left\{q_1+q_2+1\right\}\quad \because r_1+r_2=c\] \[\therefore c^{n-1}=\left\{q_1+q_2+\left(\frac{r_1+r_2}{c}\right)\right\} \] \[\therefore c^{n-1}<\left\{q_1+q_2+\left(\frac{2c}{c}\right)\right\}\quad \because r_1+r_2<2c \] \[\therefore c^{n-1}=\left\{q_1+q_2+1\right\}\quad \because \left(\frac{r_1+r_2}{c}\right)=c\] \[\therefore c^{n-1}-1=\left\{q_1+q_2\right\}\] \[\therefore \left(c^\frac{{n-1}}{2}-1\right)\left(c^\frac{{n-1}}{2}+1\right)=\left\{q_1+q_2\right\}\] \(2<n\)} \[\because c^2<a^2 <b^2<c^{n}\] \[a^2=c^{2}q_{1_2}+r_{1_2}\quad \textit{by Euclid's Division Lemma}\quad 0\leqslant r_{1_2}<c\] \[b^2=c^{2}q_{2_2}+r_{2_2}\quad 0\leqslant r_{2_2}<c\] \[a^2+b^2=c^{2}q_{1_1}+r_{1_1}+c^{2}q_{2_2}+r_{2_2}\] \[a^2+b^2=c^2\left\{q_{1_2}+q_{2_2}+\left(\frac{r_{1_2}+r_{2_2}}{c^2}\right)\right\}\] \[c^n=c^2\left\{q_{1_k}+q_{2_k}+1\right\}\] \[\therefore c^{n-2}=\left\{q_{1_2}+q_{2_2}+1\right\}\] \[\therefore (r_{1_2}+r_{2_2})=c^2\] This margin is too small for the proof