We can easily see that in the first case $a$ must be odd, in the second - even. a) $\Big ((a^2-3)^3+1\Big) ^a-1=k^2$ Then $\Big ((a^2-2)((a^2-3)^2-(a^2-3)+1)\Big) ^a=k^2+1$. Notice that $a^2-2 \equiv 3 \pmod 4$, but by Fermat's Two Squares theorem $k^2+1$ isn't divisible by a prime $3 \pmod 4$, thus no such $a$ exists. b) We can easily notice that $\Big ((a^2-3)^3+1\Big) ^{a+1}-1 \equiv 3 \pmod 4$, however, $3$ isn't a quadratic residue $\pmod 4$, so no such $a$ exists.

We can easily see that in the first case $a$ must be odd, in the second - even. a) $\Big ((a^2-3)^3+1\Big) ^a-1=k^2$ Then $\Big ((a^2-2)((a^2-3)^2-(a^2-3)+1)\Big) ^a=k^2+1$. Notice that $a^2-2 \equiv 3 \pmod 4$, but by Fermat's Two Squares theorem $k^2+1$ isn't divisible by a prime $3 \pmod 4$, thus no such $a$ exists. b) We can easily notice that $\Big ((a^2-3)^3+1\Big) ^{a+1}-1 \equiv 3 \pmod 4$, however, $3$ isn't a quadratic residue $\pmod 4$, so no such $a$ exists.