Statement The identity (n+1)3−n3=6Tn+1(n+1)^3-n^3=6T_n+1(n+1)3−n3=6Tn+1 holds. Compute D=133−123D=13^3-12^3D=133−123 and check that D≡1(mod6)D\equiv 1\pmod 6D≡1(mod6). Solution D=133−123=2197−1728=469.D=13^3-12^3=2197-1728=\boxed{469}.D=133−123=2197−1728=469. Since 469=6⋅78+1469=6\cdot 78+1469=6⋅78+1 we get D≡1(mod6)D\equiv\boxed{1}\pmod 6D≡1(mod6). Indeed 6T12+1=6⋅78+1=4696T_{12}+1=6\cdot 78+1=4696T12+1=6⋅78+1=469, matching the identity.