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Concursul interjudetean de matematica ''Marian Tarina'' p. I
Posted: Sun Jan 04, 2009 1:18 am
by alex2008
Sa se arate ca pentru orice \( n\in \mathb{N}* \) numarul \( 25^n \) se poate scrie ca suma de doua patrate perfecte .
Indicatie : \( 25^n=25^{n-1}\cdot 25 \)
Posted: Fri Apr 17, 2009 1:52 pm
by Andi Brojbeanu
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 \( n\in\mathb{N}* \).
Posted: Fri Apr 17, 2009 2:07 pm
by BogdanCNFB
\( 25^n=25\cdot 25^{n-1}=(9+16)\cdot 25^{n-1}=(3\cdot 5^{n-1})^2+(4\cdot 5^{n-1})^2 \).