Problem:
Prove that the system
x6+x3+x3y+yx3+x3y+y2+y+z9β=147157=157147β
has no solutions in integers x,y, and z.
Solution:
First Solution. Add the two equations, then add 1 to each side to obtain
(x3+y+1)2+z9=147157+157147+1(1)
We prove that the two sides of this expression cannot be congruent modulo 19. We choose 19 because the least common multiple of the exponents 2 and 9 is 18 , and by Fermat's Theorem, a18β‘1(mod19) when a is not a multiple of 19 . In particular, (z9)2β‘0 or 1 (mod19), and it follows that the possible remainders when z9 is divided by 19 are
β1,0,1(2)
Next calculate n2 modulo 19 for n=0,1,β¦,9 to see that the possible residues modulo 19 are
β8,β3,β2,0,1,4,5,6,7,9(3)
Finally, apply Fermat's Theorem to see that
147157+157147+1β‘14(mod19)
Because we cannot obtain 14 (or -5 ) by adding a number from list (2) to a number from list (3), it follows that the left side of (1) cannot be congruent to 14 modulo 19. Thus the system has no solution in integers x,y,z.
Second Solution. We will show there is no solution to the system modulo 13. Add the two equations and add 1 to obtain
(x3+y+1)2+z9=147157+157147+1
By Fermat's Theorem, a12β‘1(mod13) when a is not a multiple of 13 . Hence we compute 147157β‘41β‘4(mod13) and 157147β‘13β‘1(mod13). Thus
(x3+y+1)2+z9β‘6(mod13)
The cubes mod 13 are 0,Β±1, and Β±5. Writing the first equation as
(x3+1)(x3+y)β‘4(mod13)
we see that there is no solution in case x3β‘β1(mod13) and for x3 congruent to 0,1,5,β5 (mod13), correspondingly x3+y must be congruent to 4,2,5,β1. Hence
(x3+y+1)2β‘12,9,10, or 0(mod13)
Also z9 is a cube, hence z9 must be 0,1,5,8, or 12(mod13). It is easy to check that 6 (mod 13) is not obtained by adding one of 0,9,10,12 to one of 0,1,5,8,12. Hence the system has no solutions in integers.
Note. This argument shows there is no solution even if z9 is replaced by z3.
This problem was proposed by RΔzvan Gelca.
The problems on this page are the property of the MAA's American Mathematics Competitions