# 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 :(

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that is ${(1975^4)}^{30}$ or $1975^{(4^{30})}$ ? – Aang Sep 5 '12 at 12:49
Hint for $x,y,z\geq 0$: Which numbers are odd, which are even? – Gregor Bruns Sep 5 '12 at 12:50
Think about divisibility by $5$. – fretty Sep 5 '12 at 12:59
Avatar: $1975^{4^{30}}$ like this :x GregorBruns: They don't meantion :( – Vchau_VN Sep 5 '12 at 13:08
@Avatar: The standard grouping is $1975^{(4^30)}$ as $(1975^4)^{30}=1975^{(4\cdot 30)}$ – Ross Millikan Sep 5 '12 at 13:10

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.

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OK but how is these fraction smaller than 1/2 ? I don't get the 1/2 part – Vchau_VN Sep 5 '12 at 14:08
@Vchau_VN: Please see my edit. Does that help? – Gregor Bruns Sep 5 '12 at 14:20
so much :) thank you – Vchau_VN Sep 5 '12 at 14:32

$19^x + 5^y =- 1980z + 1975^{4^{30}}+ 2010$ 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 $2\pmod {10}$

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I just get in a high school in VN, what is RHS, LHS or mod10 @@ – Vchau_VN Sep 5 '12 at 13:10
Right Hand Side, Left Hand Side, modulo 10 (see modular arithmetic) – The Chaz 2.0 Sep 5 '12 at 13:12
RHS is the Right Hand side of the equation, LHS is the left one. $\pmod{10}$ means remainder when divided by 10. – lab bhattacharjee Sep 5 '12 at 13:14