Olimpiada Ungaria
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Olimpiada Ungaria
Aratati ca ecuatia \( (x+1)^2+(x+2)^2+...+(x+99)^2=y^z \) nu are solutii in numere intregi , \( x,y,z \) cu \( z>1 \) .
. A snake that slithers on the ground can only dream of flying through the air.
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Beniamin Bogosel
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\( \sum_{i=1}^99 (x+i)^2=99x^2+9900x+\frac{99\cdot 100 \cdot 199}{6} \).
Cred ca ideea ar fi sa demonstram ca numarul dat se divide cu 11 (ceea ce e evident) si nu se divide cu 121=\( 11^2 \).
Cred ca ideea ar fi sa demonstram ca numarul dat se divide cu 11 (ceea ce e evident) si nu se divide cu 121=\( 11^2 \).
Yesterday is history,
Tomorow is a mistery,
But today is a gift.
That's why it's called present.
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Tomorow is a mistery,
But today is a gift.
That's why it's called present.
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maxim bogdan
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Solutie
Se obtine \( y^z=33(3x^2+300x+50\cdot 199)\Longrightarrow 3|y^z. \) Cum \( z\geq 2\Longrightarrow 3^2|y^z \) Dar este evident faptul ca \( 3^2 \) nu divide \( 33(3x^2+300x+50\cdot 199). \)
E o ecuatie diofantiana cunoscuta (Hungary 1998). Apare in cartea lui Titu Andreescu: Introducere in studiul ecuatiilor diofantiene, PEN (vezi H 78 ), etc
E o ecuatie diofantiana cunoscuta (Hungary 1998). Apare in cartea lui Titu Andreescu: Introducere in studiul ecuatiilor diofantiene, PEN (vezi H 78 ), etc
Feuerbach