Prove that there is no value of the integers $x,y,z$ satisfied the equation: $19^x + 5^y + 1980z = 1975^{4^{30}} + 2010$ Prove that there is no value of the integers $x,y,z$ satisfying the equation: $$ 19^x + 5^y + 1980z = 1975^{4^{30}}+ 2010 $$
The equation $1975^{4^{30}}$ is like a double exponent :(
thanks again, it's a little hard to translate into english when my english not so well :( 
 A: Consider first the case $x,y<0$. Subtract $1980z$ from both sides of the equation. Since
$$1975^{4^{30}}+2010-1980z$$
is an integer, $19^x+5^y$ has to be as well. But for $x$ and $y$ negative, these are fractions, each of them smaller than $1/2$ since $19>2$ and $5>2$:
$$0<19^x + 5^y = \frac{1}{19^{-x}}+\frac{1}{5^{-y}}\leq \frac{1}{19}+\frac{1}{5} < 1$$
So it is impossible that this is an integer.
Now consider $x,y\geq 0$. Since $19$ is odd, $19^x$ is odd as well. The same holds for $5^y$, while $1980z$ is even for all $z$.
So the left hand side of the equation is odd + odd + even, which is even.
Can you do the same for the right hand side? You will get a contradiction.
A: $ 19^x + 5^y =- 1980z + 1975^{4^{30}}+ 2010 $ an integer.
If, at least one of $x,y$ is $<0$,the LHS=$(19^x + 5^y)$ is a fraction.
If $x,y≥0$,
$ 19^x + 5^y + 1980z = 1975^{4^{30}}+ 2010 $
$\implies 19^x=1975^{4^{30}}+ 2010-1980z-5^y$
Observe that $(10a+5)^n$ leaves remainder $5$  when divided by $10$ ,where $n$ is positive integer, $a$ is non-negative integer.
The RHS  is divisible by $5$ if $y>0$, but $(5,19)=1$, so $5∤19^x$(the LHS).
If $y=0, 19^x + 1= 1975^{4^{30}}+ 2010-1980z≡5\pmod {10} $ 
But $19^x≡±1\pmod {10}\implies 19^x + 1≡0$ or $2\pmod {10}$
