since p is prime then we need
12x+5=p^a and 12y+7=p^b
with a,b>=0 and a+b=z
so it suffices to find a prime p, which can both be 5 and 7 mod 12.
Now consider p^2-1 mod 12
p^2-1=(p+1)(p-1)
if p>2 then both p+1 and p-1 are even
thus p^2-1=0 mod 4
also if p>3 then p is either 1 or 2 mod 3
thus either p+1 or p-1 is 0 mod 3
thus p^2-1=0 mod 3
thus p^2-1=0 mod 12 for p>3
thus p^2=1 mod 12
thus means that for p>3 there are only 2 distinct powers mod 12.
namely p mod 12 and 1.
Thus for p>3 it is not possible to have both a solution to 12x+5=p^a and 12y+7=p^b.
Now we just need to check p=2 and p=3
p=2
2^z is even but both left side factors are odd. Thus no solution.
p=3
3^1=3 mod 12
3^2=9 mod 12
3^3=3 mod 12
thus there is no power of 3 congruent to 5 or 7 mod 12.
Thus there are no solutions to this problem :-)
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Posted by Daniel
on 2016-10-04 06:35:52 |
solution
