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^{...
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