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Ideea problemei este de a scrie numarul prim \( 8n+3 \) sub forma \( x^2-y^2 \)
Avem \( 4(8n+1)-24n+1=8n+3 \)
Notand \( 8n+1 \) cu \( a^2 \) si \( 24n+1 \) cu \( b^2 \)
=> \( (2a-b)(2a+b)=8n+3 \) => \( 2a-b=1 \) <=> \( 32n+4=24n+1+1+2b \) <=> \( 4n+1=b \) <=> \( 16n^2+8n+1=24n+1 \) de unde
\( n=0 \) si \( n=1 \) sunt solutii,care verifica si ipoteza
Avem \( 4(8n+1)-24n+1=8n+3 \)
Notand \( 8n+1 \) cu \( a^2 \) si \( 24n+1 \) cu \( b^2 \)
=> \( (2a-b)(2a+b)=8n+3 \) => \( 2a-b=1 \) <=> \( 32n+4=24n+1+1+2b \) <=> \( 4n+1=b \) <=> \( 16n^2+8n+1=24n+1 \) de unde
\( n=0 \) si \( n=1 \) sunt solutii,care verifica si ipoteza