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

Sep 12, 2016

See explanation...

#### Explanation:

Case $\boldsymbol{{3}^{x} + 1 = {y}^{4}}$

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

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

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

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Case $\boldsymbol{{3}^{x} - 1 = {y}^{4}}$

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 $3$:

${0}^{4} + 1 \equiv 1$

${1}^{4} + 1 \equiv 2$

${2}^{4} + 1 \equiv 2$

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