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.
The first equation is
x^4+14*x*y+1 = p^4
The other two equations are isomorphic to this one, so tabulation of solutions is the same:
x y p
1 1 2
8 22 9
15 71 16
22 148 23
29 253 30
36 386 37
20 428 23
43 547 44
50 736 51
originally this extended further, but here it only shows those cases where x+y adds up to less than 1000. In all cases y was larger than x.
It's easy to see that p must be larger than x, but we can also show that y must also be no smaller than x:
x^4+14*x*y+1 = p^4
14*x*y = p^4 - x^4 - 1
y = (p^4 - x^4 - 1) / (14 * x)
Even if p is only 1 larger than x:
(2^4 - 1^4 - 1) / (14 * 1) = 1
With larger values of p and x than 2 and 1 respectively, the gap that makes y larger than x gets even larger: In the samples p and x of 9 and 8 respectively, produces y = 22; and that's the minimal difference between p and x.
But the same is true of the isomorphic cases. y>=x and z>=y and x>=z. This is possible only when all three are equal. And all three can be equal only when all three equal 1.
So (x,y,z,p,q,r) = (1,1,1,2,2,2) is the only sextuplet that fits.
clearvars,clc
for s=2:10000 % showing original limit of x+y = 10000
for x=1:s-1
y=s-x;
lhs=x^4+14*x*y+1;
p=round(lhs^(1/4));
if p^4==lhs
fprintf('%4d %6d %5d\n',x,y,p)
end
end
end
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Posted by Charlie
on 2023-07-04 08:32:33 |
solution
