Sa se arate ca pentru orice \( n\in \mathb{N}* \) numarul \( 25^n \) se poate scrie ca suma de doua patrate perfecte .
Indicatie : \( 25^n=25^{n-1}\cdot 25 \)
Concursul interjudetean de matematica ''Marian Tarina'' p. I
Concursul interjudetean de matematica ''Marian Tarina'' p. I
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Andi Brojbeanu
- Bernoulli
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Distingem 2 cazuri:
n=2k+1:\( 25^{2k+1} \)\( =25^2^k \)\( \cdot \)\( 25= \)\( 25^2^k\cdot \)\( (4^2+3^2) \)\( =(4\cdot \)\( 25^k)^2 \)\( +(3\cdot \)\( 25^k)^2 \)
n=2k: \( 25^2^k \)\( =25^2^{k-2} \)\( \cdot \)\( 25^2 \)\( =25^2^{k-2} \)\( \cdot \)\( (24^2+7^2) \)\( =(24\cdot \)\( 25^{k-1})^2 \)\( +(7\cdot \)\( 25^{k-1})^2 \), unde \( k\geq \)\( 1 \), deoarece \( n\in\mathb{N}* \).
n=2k+1:\( 25^{2k+1} \)\( =25^2^k \)\( \cdot \)\( 25= \)\( 25^2^k\cdot \)\( (4^2+3^2) \)\( =(4\cdot \)\( 25^k)^2 \)\( +(3\cdot \)\( 25^k)^2 \)
n=2k: \( 25^2^k \)\( =25^2^{k-2} \)\( \cdot \)\( 25^2 \)\( =25^2^{k-2} \)\( \cdot \)\( (24^2+7^2) \)\( =(24\cdot \)\( 25^{k-1})^2 \)\( +(7\cdot \)\( 25^{k-1})^2 \), unde \( k\geq \)\( 1 \), deoarece \( n\in\mathb{N}* \).
Last edited by Andi Brojbeanu on Sun Dec 06, 2009 2:31 pm, edited 1 time in total.
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BogdanCNFB
- Thales
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