Find all possible sextuplets (x,y,z,p,q,r) of positive integers that satisfy this system of equations:
- x4+14xy+1 = p4
- y4+14yz+1 = q4
- z4+14xz+1 = r4
Prove that there are no further valid sextuplets satsfying the given conditions.
Notes:
• Adapted from a problem appearing in Soros Math Olympiad.
• Computer program assisted solutions are welcome, but a semi-analytic solution is preferred.
Lets just look at the first equation x^4+14xy+1=p^4
Since we are working over positive integers, the smallest p could be is x+1.
So then we can construct an inequality x^4+14xy+1>=(x+1)^4
Solving for y gives us y >= (2x^2+3x+2)/7.
If x=1 then y=1.
If x=2 then y=16/7, which is discarded as being a non-integer.
If x=3 then y=29/7, which is discarded as being a non-integer.
If x>=4 then 2x/7>1 and then 2x^2/7 > x, which then implies for all x>=4 we have y>x.
At this point we can say that x!=2 and x!=3
This analysis applies identically to the other two equations. So all the following are true:
If x=1 then y=1.
If y=1 then z=1.
If z=1 then x=1.
If x>=4 then y>x
If y>=4 then z>y
If z>=4 then x>z
If x>=4 then y>x means we can say y>=4; and then z>=4 similarly. But then y>x, z>y and x>z all must hold, forming a compound inequality x>y>z>x. Taks the extreme ends and make x>x. This is a contradiction. So there are no solutions with x>=4.
So then if x=1 then y=1 and z=1 follow, then x^4+14xy+1 = 16 = 2^4 implies p=2. q=2 and r=2 are the same. Then the one solution is (x,y,z,p,q,r) = (1,1,1,2,2,2).
Analytic Solution
