Diophantine Equation--10th Powers

The 2-1 equation

$$A^{10}+B^{10}=C^{10}$$ (1)

is a special case of Fermat's Last Theorem with $n=10$, and so has no solution. The smallest values for which $n$-1, $n$-2, etc., have solutions are 23, 19, 24, 23, 16, 27, and 7, corresponding to

$$5\cdot 1^{10}+2^{10}+3^{10}+6^{10}+6\cdot 7^{10}+4\cdot 9^{10}+10^{10}+2\cdot 12^{10}+13^{10}+14^{10}=15^{10}$$ (2)

$$5\cdot 2^{10}+5^{10}+6^{10}+10^{10}+6\cdot 11^{10}+2\cdot 12^{10}+3\cdot 15^{10}=9^{10}+17^{10}$$ (3)

$$1^{10}+2^{10}+3^{10}+10\cdot 4^{10}+7^{10}+7\cdot 8^{10}+10^{10}+12^{10}+16^{10}=11^{10}+2\cdot 15^{10}$$ (4)

$5\cdot 1^{10}+2\cdot 2^{10}+2\cdot 3^{10}+4^{10}+4\cdot 6^{10}+3\cdot 7^{10}+8^{10}+2\cdot 10^{10}+2\cdot 14^{10}+15^{10}$
$=3\cdot 11^{10}+16^{10}\quad$	(5)

$4\cdot 1^{10}+2^{10}+2\cdot 4^{10}+6^{10}+2\cdot 12^{10}+5\cdot 13^{10}+15^{10}$
$=2\cdot 3^{10}+8^{10}+14^{10}+16^{10}\quad$	(6)

$1^{10}+4\cdot 3^{10}+2\cdot 4^{10}+2\cdot 5^{10}+7\cdot 6^{10}+9\cdot 7^{10}+10^{10}+13^{10}$
$=2\cdot 2^{10}+8^{10}+11^{10}+2\cdot 12^{10}\quad$	(7)

$1^{10}+28^{10}+31^{10}+32^{10}+55^{10}+61^{10}+68^{10}$
$=17^{10}+20^{10}+23^{10}+44^{10}+49^{10}+64^{10}+67^{10}\quad$	(8)

(Lander et al. 1967).

References

Lander, L. J.; Parkin, T. R.; and Selfridge, J. L. ``A Survey of Equal Sums of Like Powers.'' Math. Comput. 21, 446-459, 1967.