Let $\mathbb{N}$ denote the set of nonnegative integers. Find all solutions $(q, r, w) \in \mathbb{N}^3$ to the equation $$2013^q+2014^w=2015^r.$$
Can someone explain me this solution ? one can post other solutions too..
$2013^q + 2014^w = 2015^r$.
Let us say that $(q, r, w) \in \mathbb{N}^3$.
We see that $r \geq 2$.
Hence, $\min\{ \nu _{31}(LHS)\} = 2$.
We see that $2013 \equiv -2 \pmod{961}$ and $2014 \equiv -1 \pmod{961}$.
Hence, we require $2013^q \equiv 960$ or $1 \pmod{961}$.
But, we see by the cycle of modulo of powers of $-2$ $\pmod{961}$ that this condition is never satisfied.
Hence, we get that the only triplet is $(q, w, r) = (0, 1, 1)$.
Actually , this is a pretty fun problem , which plays with modulos so much .
Solution: $\boxed{(q, w, r) = (0, 1, 1)}$
Note that $r>0 $ .
Also, note that $w>0$, (if not then $2013^q$ is odd, $2014^w=1$ and $2015^r$ is odd, not possible as RHS is even and LHS is odd )
Claim: $q=0$
Proof: let's say for contradiction , assume that q /ge 1
then $2013^q \equiv 0 \pmod 3$
and $2014^w \equiv 1 \pmod 3$
So we have $2013^q + 2014^w \equiv 1 mod 3 \implies r$ is even.
Let $r=2b$ .
Now, let us look at the unit digits.
we see that $2013^q$ ends with $3,9,7,1$ and $2014^w$ ends with $4,6$ and $2015^r$ ends with $5$.
By calculations, we get that q must be even , hence let $q=2a$.
Hence we have $$ 2013^{2a} + 2014^w =2015^{2b} \implies 2014^w=2015^{2b}-2013^{2a}\implies 2014^w= (2015^b-2103^a)(2015^b+2103^a) \implies 2^w \cdot 19^w \cdot 53^w = (2015^b-2103^a)(2015^b+2103^a)$$
Let $\alpha=(2015^b+2103^a)$ and $\beta=(2015^b-2103^a)$.
Clearly $\alpha>\beta$.
let $d=\text{gcd($\alpha,\beta$)}$
Note that $d\mid 2014^w$ and $d\mid (\alpha +\beta)=2\cdot 2015^b \implies d=1$ or $2$ .
Now, since $\alpha>\beta$ , and $d= 1$ or $2$, we get that $v_{53}(\alpha)=w$ and $v_{19}(\beta)=w $.
Now, in $\alpha , 2015^b \equiv 1 \pmod {53}$ , since $2015 \equiv 1 \pmod {53} $.
Since $2015^b - 2013^a \equiv 0 \pmod 53 \implies a$ is odd .
In $\beta, 2015^b \equiv 1 \pmod {19}$ , since $2015 \equiv 1 \pmod {19}$ .
Now, since $a$ is odd $ \implies 2013^a \equiv -1 \pmod {19} \implies \beta \equiv 2 \pmod {19} \implies v_{19}(\beta)=w=0$ , a contradiction, since in the first observation,we noted that $w>0$ .
Hence $q=0$
Main Proof:
So, we have $1+2014^w=2015^r$ . Note that $r$ is odd since $1 \equiv 1 \pmod 3, 2014\equiv 1 \pmod 3 $.
hence $2014^w=2015^r-1=2014(2015^{r-1} +\dots +1)$ .
But note $2015^{r-1} +\dots +1$ ( since $r$ is odd) .
Hence $w=1$ .
Hence $2013^q+2014^w=2013^0+2014^1= 1+2014 =2015 \implies r=1$.
And we are done!I hope someone can verify this solution .