 # Prove that x^n_1+x^n_2 is an integer and is not divisible by 5 saucletbh 2022-09-18 Answered
Prove that ${x}_{1}^{n}+{x}_{2}^{n}$ is an integer and is not divisible by 5
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We have . From the equation, we have

Adding both we get
${x}_{1}^{2}+{x}_{2}^{2}=6\underset{\text{Integer}}{\underset{⏟}{\left({x}_{1}+{x}_{2}\right)}}-2=\text{Integer}$
Now use strong induction and make use of the fact that

i.e.,
${x}_{1}^{n+2}+{x}_{2}^{n+2}=6\left({x}_{1}^{n+1}+{x}_{2}^{n+1}\right)-\left({x}_{1}^{n}+{x}_{2}^{n}\right)$
Use the same idea to show that
${x}_{1}^{n}+{x}_{2}^{n}\equiv \left\{\begin{array}{ll}1\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}5\right)& n\equiv 1\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}4\right)\\ 4\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}5\right)& n\equiv 0,2\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}4\right)\\ 3\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}5\right)& n\equiv 3\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}4\right)\end{array}$

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