Exercise 8. Find all integers a, b, c $\in \mathbb{N}$ such that $1517^{\mathrm{a}}+15^{\mathrm{b}}=1532^{\mathrm{c}}$.
Solution to Exercise 8 We start by handling the small values of $c$.

If $c=0, 1517^{a}+15^{b}=1$ has no solution.

If $c=1$, we need to find $a, b$ such that $1517^{a}+15^{b}=1532$. If $a \geqslant 2$ or if $a=0$ there is no solution. If $a=1$, then $15^{b}=1532-1517=15$ so $b=1$: this provides the solution $(1,1,1)$.

Now assume $c \geqslant 2$. In particular, $16=4^{2}$ divides $1532^{\circ}$: this invites us to study the equation modulo powers of 2 less than 16 to eliminate the dependence on $c$. We then get $1+(-1)^{b} \equiv 0(\bmod 4)$ so $b$ is odd (in particular, $b \geqslant 1$).

By reducing modulo $8: 15^{b} \equiv-1(\bmod 8)$ (since $b$ is odd). We then have $1517^{a}-1 \equiv 5^{a}-1 \equiv 0(\bmod 8)$, so $a$ is even.

By reducing modulo 5 (since 5 divides 15, this eliminates the dependence on $b$), we have $2^{a} \equiv 2^{c}(\bmod 5)$, so $a \equiv c(\bmod 4)$ (the order of 2 modulo 5 is 4). Since we have shown that $a$ is even, $c$ is even. Let $a=2e$ and $c=2d$, then $d$ and $e$ have the same parity (since $2e \equiv 2d(\bmod 4)$ so $e \equiv d(\bmod 2)$), and by substituting back into the original equation, we get $15^{b}=\left(1532^{d}-1517^{e}\right)\left(1532^{d}+1517^{e}\right)$.

Notice then that

$$
\left(1532^{d}+1517^{e}\right)-\left(1532^{d}-1517^{e}\right)=2 \times 1517^{e}=2 \times 37^{e} \times 41^{e}
$$

Neither 37 nor 41 divides 1532, so $1532^{d}-1517^{e}$, and $1532^{d}-1517^{e}$ is odd. Therefore, $1532^{d}+1517^{e}$ and $1532^{d}-1517^{e}$ are coprime, so one of them is $5^{b}$ and the other is $3^{b}$. Since $1532^{d}+1517^{e}>1532^{d}-1517^{e}$, it follows that $1532^{d}+1517^{e}=5^{b}$ and $1532^{d}-1517^{e}=3^{b}$. Then $3^{b}+5^{b}=2 \cdot 1532^{d}$. If $d \geqslant 2$, the right-hand side is divisible by 16, but we check modulo 16 that the left-hand side is never divisible by $16\left(3^{4} \equiv 5^{4} \equiv 1(\bmod 16)\right.$ so it suffices to test $3^{0}+5^{0} \equiv 2(\bmod 16), 3^{1}+5^{1} \equiv 8(\bmod 16), 3^{2}+5^{2} \equiv 2(\bmod 16)$ and $\left.3^{3}+5^{3} \equiv 5(\bmod 16)\right)$, which implies that $d=1($ since $2d=c \geqslant 2)$ and thus $e$ is odd $(d \equiv e(\bmod 2))$. Since $1532^{d}>1517^{e}$, it follows that $e=1$, but $1532^{d}-1517^{d}=15$ is not a power of 3, leading to a contradiction.

Finally, the only solution is $(a, b, c)=(1,1,1)$.

Comment from the graders: A problem not often tackled but well understood by those who submitted a solution. The goal was to find properties of $a, b$, and $c$ by reducing modulo small numbers until reaching an absurdity, and many paths were followed. However, be careful, it is generally necessary to handle small cases separately. For example, here, $15^{0}=1 \not \equiv 0[3]$!