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Victorboss21
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[tex] {2019}^{2019} [/tex]
ca o sumă de 4 pătrate perfecte​

Răspuns :

Babybianca
2019^2019 = 2019 × 2019^2018 = (1936 + 49 + 25 + 9) × 2019^2018 = (44^2 + 7^2 + 5^2 + 3^2) × 2019^2018 = 44^2 × 2019^2018 + 7^2 × 2019^2018 + 5^2 × 2019^2018 + 3^2 × 2019^2018 ÷ (44×2019^1009)^2 + (7×2019^1009)^2 + (5×2019^1009)^2 + (3×2019^1009)^2

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