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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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    @Avatar: The standard grouping is $1975^{(4^30)}$ as $(1975^4)^{30}=1975^{(4\cdot 30)}$2012-09-05

2 Answers 2

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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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$ 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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    RHS is the Right Hand side of the equation, LHS is the left one. $\pmod{10}$ means remainder when divided by 10.2012-09-05