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Solve the equation on the set of natural numbers: $25^{x+1}+4^{y-2}-2 \cdot 5^{x+2}-2^{y}+29=0$

  • 0
    See here: https://www.wolframalpha.com/input/?i=25%5E%7Bx%2B1%7D+%2B+4%5E%7By-2%7D-2%5Ccdot+5%5E%7Bx%2B2%7D-2%5Ey+%2B+29&wal=header – 2017-02-26
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    Is this actually a $8$-th-grade-exercise ? – 2017-02-26

2 Answers 2

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Suppose $x,y$ are real numbers such that

$$\qquad\; 25^{x+1}+4^{y-2}-2 \cdot 5^{x+2}-2^{y}+29=0$$

Then letting $u = 5^{x+1}$ and $v = 2^{y-2}$,

\begin{align*} &25^{x+1} + 4^{y-2} - 2 \cdot 5^{x+2} - 2^{y} + 29 = 0\\[6pt] \iff\; &u^2 + v^2 - 10u - 4v +29 = 0\\[6pt] \iff\; &(u-5)^2 + (v-2)^2 = 0\\[6pt] \iff\; &u = 5\, \text{ and } \,v = 2\\[6pt] \iff\; &x + 1 = 1\, \text{ and } \,y - 2 = 1\\[6pt] \iff\; &x = 0\, \text{ and } \,y = 3\ \end{align*}

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We have $$S:=25^{x+1}-2\cdot 5^{x+2}=5^{2x+2}-2\cdot 5^{x+2}=5^{x+2}(5^x-2)$$

and $$T:=4^{y-2}-2^y=2^{2y-4}-2^y=2^y(2^{y-4}-1)$$

We have to solve $S+T=-29$

$S$ is only negative for $x=0$. In this case we have $S=-25$

$T$ is only negative for $y< 4$. The values are $-\frac{15}{16},-\frac{7}{4},-3,-4$

We see that $x=0$ and $y=3$ is the only solution because the minimum value of $S+T$ is $-29$