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Question Number 184567 by a.lgnaoui last updated on 08/Jan/23

Resoudre dans Z^+   (√a)  +(√b) =z    (a,b,z)∈N^3   a,b ?

$${Resoudre}\:{dans}\:\mathbb{Z}^{+} \\ $$$$\sqrt{{a}}\:\:+\sqrt{{b}}\:={z}\:\:\:\:\left({a},{b},{z}\right)\in\mathbb{N}^{\mathrm{3}} \\ $$$${a},{b}\:? \\ $$

Answered by Frix last updated on 08/Jan/23

a=n_1 ^2 ∧b=n_2 ^2   ⇒  z=(√a)+(√b)=n_1 +n_2

$${a}={n}_{\mathrm{1}} ^{\mathrm{2}} \wedge{b}={n}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${z}=\sqrt{{a}}+\sqrt{{b}}={n}_{\mathrm{1}} +{n}_{\mathrm{2}} \\ $$

Commented by a.lgnaoui last updated on 08/Jan/23

n_1 ?     n_2 ?

$${n}_{\mathrm{1}} ?\:\:\:\:\:{n}_{\mathrm{2}} ? \\ $$

Commented by Frix last updated on 08/Jan/23

n_1 ∈N; n_2 ∈N

$${n}_{\mathrm{1}} \in\mathbb{N};\:{n}_{\mathrm{2}} \in\mathbb{N} \\ $$

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