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Question Number 8035 by Nayon last updated on 28/Sep/16
prove >> a^n +b^n =c^n   [n>2]  it has no integer roots
$${prove}\:>>\:{a}^{{n}} +{b}^{{n}} ={c}^{{n}} \:\:\left[{n}>\mathrm{2}\right] \\ $$$${it}\:{has}\:{no}\:{integer}\:{roots} \\ $$$$ \\ $$
Commented by FilupSmith last updated on 29/Sep/16
Do you mean:  Prove that  a^n +b^n ≠c^n     (∀n,a,b,c∈N:n>2)
$$\mathrm{Do}\:\mathrm{you}\:\mathrm{mean}: \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\:{a}^{{n}} +{b}^{{n}} \neq{c}^{{n}} \:\:\:\:\left(\forall{n},{a},{b},{c}\in\mathbb{N}:{n}>\mathrm{2}\right) \\ $$
Commented by javawithfish last updated on 29/Sep/16
ask it to andrew wilze
$${ask}\:{it}\:{to}\:{andrew}\:{wilze} \\ $$

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