Continuing from here . Book Referred: David Burton, Elementary Number Theory Also, thanks to Pranjal Bhaiya and Ritam bhaiya for taking lectures in NT in EGMO Camp and Sophie! We state and prove LTE first. $$v_p(x^n-y^n)=v_p(x-y)+v_p(n), n|x-y, n\not\mid x.$$ Walkthough: We use induction on $v_p(n).$ [ We show for $v_p(n)=0, v_p(n)=1$ and then use induction. 1. We show it for $v_p(n)=0.$ That is show $v_p(x^n-y^n)=v_p(x-y)$ To show this is true, $$v_p(\frac{x^n-y^n}{x-y})=v_p(x^{n-1}+yx^{n-2}+y^2x^{n-3}+\dots+y^{n-1})=0.$$ As $x\equiv y\pmod p.$ So, $$x^{n-1}+yx^{n-2}+y^2x^{n-3}+\dots+y^{n-1}\equiv nx^{n-1}\pmod p. $$ And $p\not\mid nx^{n-1}$ 2. We show it for $v_p(n)=1.$ That is show $v_p(x^n-y^n)=v_p(x-y)+1$ To show this is true, $$v_p(\frac{x^n-y^n}{x-y})=v_p(x^{n-1}+yx^{n-2}+y^2x^{n-3}+\dots+y^{n-1})=1.$$ As $x\equiv y\pmod p\implies x=y+pk$ So, $$x^{n-1}+yx^{n-2}+y^2x^{n-3}+\dots+y^{n-1}\pmod {p^2} $$ $$\equiv (pk+y)^{n-1}+(pk+y)^{n-2}y+(pk+y)^{n-3}y^2+\dots+y^{n-1}\pmod {p^2

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