Doubles
2005 BCR Open Romania
Champions	José Acasuso Sebastián Prieto
Runners-up	Victor Hănescu Andrei Pavel
Score	6–3, 4–6, 6–3

Lucas Arnold Ker and Mariano Hood were the defending champions, but Arnold Ker did not compete this year. Hood teamed up with Martín García and lost in the quarterfinals to Juan Ignacio Chela and Luis Horna.

José Acasuso and Sebastián Prieto won the title by defeating Victor Hănescu and Andrei Pavel 6–3, 4–6, 6–3 in the final.^[1]

Seeds

Draw

Draw

	First round							Quarterfinals							Semifinals							Final
	1	M García M Hood	6	56	6
	WC	V Ioniță R Sabău	1	68	2			1	M García M Hood	3	3
		JI Chela L Horna	6	6					JI Chela L Horna	6	6
		T Behrend A Waske	2	2												JI Chela L Horna	3	4
	3	J Acasuso S Prieto	6	6											3	J Acasuso S Prieto	6	6
		J Coetzee C Haggard	4	4				3	J Acasuso S Prieto	6	6
		F Mayer R Wassen	6	67					F Mayer R Wassen	1	2
		PH Mathieu R Rochus	4	51																		3	J Acasuso S Prieto	6	4	6
		V Hănescu A Pavel	6	67																			V Hănescu A Pavel	3	6	3
		P Pála D Škoch	4	53					V Hănescu A Pavel	6	67
		N Almagro A Martín	2	2				4	M Fyrstenberg M Matkowski	2	53
	4	M Fyrstenberg M Matkowski	6	6												V Hănescu A Pavel	6	6
	WC	I Moldovan G Moraru	6	6											WC	I Moldovan G Moraru	4	4
		C Berlocq M Puerta	1	2				WC	I Moldovan G Moraru	610	6
		D Sanguinetti T Vanhoudt	6	67					D Sanguinetti T Vanhoudt	58	4
	2	J Knowle M Kohlmann	4	50

References

External links

• Official results archive (ATP)
• Official results archive (ITF)