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 :( 

$ 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 $2\pmod {10}$
