I had lost: the target date is up! Here is farewell miniature It is common knowledge: If A^n+B^n=C^n, and ABC does not divided by prime n>2, then, according to Little Fermat’s theorem, in base n for number A, B, C there exist next equalities: 1a°. A^n=(C-B)P, где C-B=a^n, P=p^n, A=ap; 1b°. B^n=C^n-A^n=(C-A)Q, где C-A=b^n, Q=q^n, A=aq 1c°. C^n=A^n+B^n=(A+B)R, где A+B=c^n, R=r^n, A=ar and the numbers 2°. A+B-C=ap+bq-cr=u’n^k, where k>0 (corollary from Little Fermat’s theorem). 3°. (A+B)-(C-B)-(C-A)=2u’n^k, where numbers (A+B), (C-B), (C-A) are composite. Then, after transformation of the last digit in the number c into n-1 (with help of multiplication of Fermat’s equality by corresponding number d^n), from 2°-3° the next equalities follow: 4a°. a*x=(c*-b*)x, 4b°. b*y=(c*-a*)y, 4c°. c*z=(a*+b*)z where a*, b*, c* are the last digits in the numbers a, b, c, x, y, z are the last digits in the numbers p, q, r [and numbers (c^n-b^n)/(c-b), (c^n-a^n)/(c-a), (a^n+b^n)/(a+b)]. On rewrite 3° as: 5°. (a*+b*)z-(c*-b*)x-(c*-a*)y=a*z+b*z-c*x+b*x-c*y+a*y=2u’n^k, or a*(z+y)+b*(z+x)-c*(x+y)= 2u’n^k. Comparing 5° and 4°, we have: x=z+y, y=z+x, z=x+y, whence it follows that x=y=z=0. But according to Little Fermat’s theorem, x=y=z=1. Therefore, Fermat’equality has no positive solutions. If c* (or b*) is equal to zero, then also x=y=z=0. Happy New Year!