Find four positive integers P, Q, R and S that simultaneously satisfy this system of diophantine equations:

• P³ - Q³ = R⁵, and:
• P⁵ - Q⁵ = S³

Let {u,v} be whole numbers with u>v.

Then (u(u^3-v^3)^3(u^5-v^5)^10)^3 - (v(u^3-v^3)^3(u^5-v^5)^10)^3 = (u^3-v^3)^10 (u^5-v^5)^30, a 5th power;

And (u(u^3-v^3)^3(u^5-v^5)^10)^5 - (v(u^3-v^3)^3(u^5-v^5)^10)^5=(u^3-v^3)^15 (u^5-v^5)^51, a cube.

The explanation can best be demonstrated by example.

(2*x^3)^3-(x^3)^3 =7x^9, and if x=7, then (2*7^3)^3-(7^3)^3 =49^5, a 5th power, because x=7 causes the 9th power to tick over into a 10th, and hence a 5th, power, and so on.

Now for some {u,v,b} we have (u(u^3-v^3)^3b)^3-(v(u^3-v^3)^3b)^3 =b^3(u^3-v^3)^10, and b must be a 5th power at least.

But also (u(u^3-v^3)^3b)^5-(v(u^3-v^3)^3b)^5 = b^5 (u^3-v^3)^15 (u^5-v^5), which ticks over in the same way if b is the 10th power of (u^5-v^5): (b^5)^10=b^50*b=b^51.

So, for example:

P=3599205224401089975788028777     
Q=2399470149600726650525352518     
R=31856784207426700     
S=8064356992023183474087166067697520234761431929

is also a solution, with {u,v} ={3,2}, In Benny's solution, {u,v} ={2,1} hence that solution is the smallest.

Other compliant powers of b^x also suffice: if x=25, then R = (u^3-v^3)^2(u^5-v^5)^15, S = (u^3-v^3)^3(u^5-v^5)^42, and so on.

Several other methods of constructing candidates were also attempted, without success; if there is a smaller solution set I would be intrigued to see it.

Edited on November 16, 2014, 1:46 am

Posted by broll on 2014-11-15 23:55:07