Prove that $3^{x} - 1 = y^{4}$ $3^x-1=y^4$ or $3^{x} + 1 = y^{4}$ $3^x+1=y^4$ have not integer positive solutions. ?

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Prove that $3^{x} - 1 = y^{4}$ $3^x-1=y^4$ or $3^{x} + 1 = y^{4}$ $3^x+1=y^4$ have not integer positive solutions. ?

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Answer 1 · 2016-09-12T20:02:50

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Explanation:

Case $3^{x} + 1 = y^{4}$ $bb(3^x+1 = y^4)$

If $3^{x} + 1 = y^{4}$ $3^x +1 = y^4$ then:

$3^{x} = y^{4} - 1 = (y - 1) (y + 1) (y^{2} + 1)$ $3^x = y^4-1 = (y-1)(y+1)(y^2+1)$

If $y$ $y$ is an integer, then at least one of $y - 1$ $y-1$ and $y + 1$ $y+1$ is not divisible by $3$ $3$ , so they cannot both be factors of an integer power of $3$ $3$ .

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Case $bb(3^x-1 = y^4)$

If $3^x - 1 = y^4$ then:

$3^x = y^4 + 1$

Consider possible values of $y^4+1$ for the values of $y$ modulo $3$ :

$0^4 + 1 -= 1$

$1^4 + 1 -= 2$

$2^4 + 1 -= 2$

Since none of these is congruent to $0$ modulo $3$ , they can not be congruent to $3^x$ for positive integer values of $x$ .

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Prove that $3^{x} - 1 = y^{4}$ $3^x-1=y^4$ or $3^{x} + 1 = y^{4}$ $3^x+1=y^4$ have not integer positive solutions. ?

1 Answer

Explanation:

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Prove that $3^{x} - 1 = y^{4}$ $3^x-1=y^4$ or $3^{x} + 1 = y^{4}$ $3^x+1=y^4$ have not integer positive solutions. ?