Obviously left hand and right sides of the equality congruent to $2^a$ and $2^c$ modulo 15, respectively. Hence $a=c$. By modulo 7 we can esily obtain that $2^a+1=3^c=3^a$. Since $2+1$ divedes $3$ by LTE we have $v_3 (2^a+1)=v_3 a+1$. If $a=3t$ then $27^t=8^t +1$, clearly $27^t >8^t +1$. Therefore 3 does not divide $a$ which shows that $a=1$ by this $b=1$. So the answer is$(a;b;c)=(1;1;1)$.

$2^a \equiv 2^c \pmod{15}$ gives $a \equiv c \pmod{4}$ and not $a=c$.

Obviously $a,b,c>0$(it's easy by Mihailescu's Theorem). We have $0\equiv 2012^c\equiv 1997^a+15^b\equiv 1+(-1)^b$,which implie $b\equiv 1(mod 2)$. Now suppose that $c\ge 2$.Then $0\equiv 2012^c\equiv 1997^a+15^b\equiv 1997^a+7(mod 8)$,hence $1997^a\equiv 1(mod 8)$,which implies $a\equiv 0(mod 2)$. Now $1997^a+15^b\equiv 2^a(mod 5)$ and $1997^a+15^b\equiv 2012^c\equiv 2^c(mod 5)$,so $2^a\equiv 2^c(mod 5)$.This implies $a\equiv c(mod 4)$ i.e. $0\equiv a\equiv c(mod 2)$. Now just factor $15^b=(2012^{\frac{c}{2}}-1997^{\frac{a}{2}})(2012^{\frac{c}{2}}+1997^{\frac{a}{2}})$.Now the rest is easy.

huricane wrote: Now just factor $15^b=(2012^{\frac{c}{2}}-1997^{\frac{a}{2}})(2012^{\frac{c}{2}}+1997^{\frac{a}{2}})$.Now the rest is easy. Can you post your solution for the case $2012^{c/2}-1997^{a/2}=3^b$ and $2012^{c/2}-1997^{a/2}=5^b$ ?

Of course I can show my solution.If $b\equiv 0(mod 2)$,then $2012^{\frac{c}{2}}-1997^{\frac{a}{2}}=3^b\equiv 1(mod 4)$ and $2012^{\frac{c}{2}}+1997^{\frac{a}{2}}=5^b\equiv 1(mod 4)$,so $2\cdot 1997^{\frac{a}{2}}\equiv 0(mod 4)$,contradiction!Thus $b\equiv 1(mod 2)$.Now looking at the equations modulo $3$ it can be easily seen that $\frac{a}{2}\equiv \frac{c}{2}\equiv 0(mod 2)$.Now just factor $2012^{\frac{c}{2}}-1997^{\frac{a}{2}}=(2012^{\frac{c}{4}}-1997^{\frac{a}{4}})(2012^{\frac{c}{4}}+1997^{\frac{a}{4}})=3^b$.The rest is easy.

huricane wrote: Obviously $a,b,c>0$(it's easy by Mihailescu's Theorem). Or because the problem says so...

zachman99323 wrote: huricane wrote: Obviously $a,b,c>0$(it's easy by Mihailescu's Theorem). Or because the problem says so... The problem doesn't say so,$\mathbb{N}=\{0,1,2,3...\}$.Just take a look here,for example:http://en.wikipedia.org/wiki/Natural_number .

huricane wrote: zachman99323 wrote: huricane wrote: Obviously $a,b,c>0$(it's easy by Mihailescu's Theorem). Or because the problem says so... The problem doesn't say so,$\mathbb{N}=\{0,1,2,3...\}$.Just take a look here,for example:http://en.wikipedia.org/wiki/Natural_number . "There is no universal agreement about whether to include zero in the set of natural numbers." <-- says that on the wikipedia page, but since the problem doesn't define $\mathbb{N}$, I guess you should prove that none of them are 0.

We can kill problame with mod(16).

HECAM-CA-CEBEPA wrote: We can kill problame with mod(16). In details: Suppose $c=>2$. Then $LHS =(-3)^a+(-1)^b=0(mod 16)$. $(-3)^a=-3, 9, 5, 1$ and $(-1)^b=1, -1$. Checking all the possibilities, we find that the only possibility is $a=0(mod4)$ and $b=1(mod2)$ By modulo 5 see that $c$ is even. Then continue as the solution in post #4 and #6

Answer:$(1,1,1)$ By $mod 7$ we get \[2^a+1 \equiv 3^c (mod 7)\]$\implies 3$ doesn't divide $c$. If $b \geq 2$ then. \[(-1)^a \equiv 5^c (mod 9) \]$\implies 3|c$ Contradiction So $b=1$. We have $1997^a+15=2012^c$ $a=b=c=1$ is a solution. Let $c \geq 2 $ then $a,c \geq 2$ $5^a-1 \equiv 0 (mod 8) \implies a$ is even. $a=2k$ $7 \equiv 5^c (mod 7) \implies c$ is even. $c=2m$ $15=(2012^m-1997^k)(2012^m+1997^k)$ $i)2012^m-1997^k=1,2012^m+1997^k=15$ No solution $ii)2012^m-1997^k=3,2012^m+1997^k=5$ No solution