a^{2}+b^{2}=c^{2}and ab/2=2008 Cool!!! 2008 is a congruent number! Substitute n=2008 to y^{2}=x^{3}-n^{2}x and solve about (x,y),then we obtain (x,y)=(-33189589138608236/173526479973025,-63190422991165235540565912/2285854776728259808625) a=(n^{2}-x^{2})/y=10946706347066226/75738545443585 b=2nx/y=605908363548680/21806187942363 c=(n^{2}+x^{2})/y=243077092075274283595010477962/1651568956424015360248091355 a^{2}+b^{2}=c^{2}and ab/2=2008 When I used the Tunnell's Theorem,I noticed a certain pattern. A certain pattern is ,if n = 5,6,7 mod 8, then n becomes a congruent number. I noticed that the corresponding equations always have no solutions. Therefore, such number n = 5,6,7 mod 8 is always a congruent number by Tunnell's Theorem. I proved this pattern which I noticed as a next lemma. Consequently,if we check only whether n is congruent number or not by Tunnell's theorem, we can exclude any natural number n=5,6,7 mod 8 from cheking.Tunnell's Theorem Let n be a squarefree natural number. If n is odd then n is congruent number,then the number of solutions of x^{2}+2y^{2}+8z^{2}=n is equal to the number of solutions of x^{2}+2y^{2}+32z^{2}=n. If n is even then n is congruent number,then the number of solutions of x^{2}+4y^{2}+8z^{2}=n/2 is equal to the number of solutions of x^{2}+4y^{2}+32z^{2}=n/2. Furthermore,if the Birch and Swinnerton-Dyer conjecture holds,then the converse also holds.LemmaAny natural number n=5,6,7 mod 8 is congruent number if the Birch and Swinnerton-Dyer conjecture holds.Proof.(1). Squarefree natural number n=5,7 mod 8 is congruent number.if n is squarefree natural number and odd,examine that the number of solutions of x^{2}+2y^{2}+8z^{2}=n and number of solutions of x^{2}+2y^{2}+32z^{2}=n. x^{2}={0,1,4} mod 8,so n=x^{2}+2y^{2}={0,1,2,3,4,6} mod 8 So,if n=5,7 mod 8 then x^{2}+2y^{2}+8z^{2}=n and x^{2}+2y^{2}+32z^{2}=n has no solution. Therefore,the number of solutions of x^{2}+2y^{2}+8z^{2}=n is equal to the number of solutions of x^{2}+2y^{2}+32z^{2}=n. By Tunnell' theorem,squarefree natural number n=5,7 mod 8 is congruent number.(2). Squarefree natural number n=6 mod 8 is congruent number. if n is squarefree natural number and even,examine that the number of solutions of x^{2}+4y^{2}+8z^{2}=n/2 and number of solutions of x^{2}+4y^{2}+32z^{2}=n/2. In the same way as the odd case,n/2=x^{2}+4y^{2}={0,1,4,5} mod 8 So,if n=6 mod 8 x^{2}+4y^{2}+8z^{2}=n/2 and x^{2}+4y^{2}+32z^{2}=n/2 has no solution. Therefore,the number of solutions of x^{2}+4y^{2}+8z^{2}=n/2 is equal to the number of solutions of x^{2}+4y^{2}+32z^{2}=n/2. By Tunnell' theorem,squarefree natural number n=6 mod 8 is congruent number.(3). If natural number dif d^{2}n=5,6,7 mod 8 then n must be 5,6,7 mod 8.^{2}n=5 mod 8 then d =1 mod 8,so n=5 mod 8. In the same way ,n=5,6,7 mod 8. Consequently, by (1),(2),and (3), Any natural number n=5,6,7 mod 8 is congruent number.

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