Vertical shift of polynomials with integer roots
The roots of such polynomial pairs are essentially the ideal solutions of the Prouhet–Tarry–Escott problem .
According to above wiki entry, ideals solutions are known for $3 \le n \le 10$ and for $n = 12$. No ideal solution is known for $n = 11$ or for $n \ge 13$.
Following solution for $n = 12$ is extracted from Chen Shuwen's Equal sums of like powers page. Let $$\begin{align} (a)_{i=1}^{12} &= (1,12,25,66,91,130,174,213,238,279,292,303)\\ (b)_{i=1}^{12} &= (4,6,31,58,105,117,187,199,246,273,298,300) \end{align} $$ and consider following two polynomials of degree $12$, $$A(x) = \prod_{i=1}^{12} (x - a_i)\quad\text{ and }\quad B(x) = \prod_{i=1}^{12} (x - b_i)$$ We have $$B(x) - A(x) = 67440294559676054016000$$
I don't expect there will be an upper bound.
In the C++ program at the end, note how s4 and t4 are ignored with 4 integers.
The first bit is gp-Pari
parisize = 4000000, primelimit = 500000
? (x-1)*(x-5)*(x-8)*(x-12)
%1 = x^4 - 26*x^3 + 221*x^2 - 676*x + 480
? (x-2)*(x-3)*(x-10)*(x-11)
%2 = x^4 - 26*x^3 + 221*x^2 - 676*x + 660
?
====================================================
jagy@phobeusjunior:~$ ./mse
1 5 8 12 2 3 10 11
2 6 9 13 3 4 11 12
1 7 8 14 2 4 11 13
3 7 10 14 4 5 12 13
2 8 9 15 3 5 12 14
4 8 11 15 5 6 13 14
3 9 10 16 4 6 13 15
1 6 11 16 2 4 13 15
5 9 12 16 6 7 14 15
1 8 10 17 2 5 13 16
4 10 11 17 5 7 14 16
2 7 12 17 3 5 14 16
6 10 13 17 7 8 15 16
1 9 10 18 3 4 15 16
2 9 11 18 3 6 14 17
5 11 12 18 6 8 15 17
3 8 13 18 4 6 15 17
7 11 14 18 8 9 16 17
2 10 11 19 4 5 16 17
1 8 12 19 3 4 16 17
3 10 12 19 4 7 15 18
====================================================
int main()
{
for(mpz_class d = 4; d <= 25; ++d){
for(mpz_class c = 3; c < d; ++c){
for(mpz_class b = 2; b < c; ++b){
for(mpz_class a = 1; a < b; ++a){
mpz_class s1,s2,s3,s4;
s1 = a + b + c + d ;
s2 = a*a + b*b + c*c + d*d;
s3 = a*a*a + b*b*b + c*c*c + d*d*d;
s4 = a*a*a*a + b*b*b*b + c*c*c*c + d*d*d*d;
for(mpz_class h = 4; h < d; ++h){
for(mpz_class g = 3; g < h; ++g){
for(mpz_class f = 2; f < g; ++f){
for(mpz_class e = 1; e < f; ++e){
mpz_class t1,t2,t3,t4;
t1 = e + f + g + h ;
t2 = e*e + f*f + g*g + h*h;
t3 = e*e*e + f*f*f + g*g*g + h*h*h;
t4 = e*e*e*e + f*f*f*f + g*g*g*g + h*h*h*h;
if( s1 == t1 && s2 == t2 && s3 == t3 )
{
cout << setw(4) << a << setw(4) << b << setw(4) << c << setw(4) << d;
cout << " ";
cout << setw(4) << e << setw(4) << f << setw(4) << g << setw(4) << h;
cout << endl;
}
}}}} // efgh
}}}} // abcd
return 0;
}
=================================================
here is quintic, all positive integer roots (distinct) with the smallest maximal element, which turns out to be 19.
2 3 11 15 19 1 5 9 17 18
===========================================================
? ( x-2)*(x-3)*(x-11)*(x-15)*(x-19)
%3 = x^5 - 50*x^4 + 890*x^3 - 6700*x^2 + 19629*x - 18810
? ( x-1)*(x-5)*(x-9)*(x-17)*(x-18)
%4 = x^5 - 50*x^4 + 890*x^3 - 6700*x^2 + 19629*x - 13770
?
==========================================================
quintic program slowed way, way, down, I had it print the time at each success
jagy@phobeusjunior:~$ ./mse
Sat Aug 24 16:46:03 PDT 2019
progress 10
2 3 11 15 19 1 5 9 17 18
Sat Aug 24 16:50:39 PDT 2019
progress 20
3 4 12 16 20 2 6 10 18 19
Sat Aug 24 16:54:37 PDT 2019
2 4 13 15 21 1 7 9 18 20
Sat Aug 24 16:59:55 PDT 2019
4 5 13 17 21 3 7 11 19 20
Sat Aug 24 17:01:26 PDT 2019
3 5 14 16 22 2 8 10 19 21
Sat Aug 24 17:10:22 PDT 2019
5 6 14 18 22 4 8 12 20 21
Sat Aug 24 17:12:45 PDT 2019
4 6 15 17 23 3 9 11 20 22
Sat Aug 24 17:27:05 PDT 2019
6 7 15 19 23 5 9 13 21 22
Sat Aug 24 17:30:53 PDT 2019