30 and 60 from contestcen.com/digits.html

 30 and 60
       Find the smallest 30-digit integer N which contains each of the digits 0 to 9 three times, whose square power is a 60-digit number which contains each of the digits 0 to 9 six times.

import time
time1=time.time()

''' This part should be changed for the problem.
stri,globalstop='31622',True

repeatcount,totalcount,allcount,mulcount=4,8,40,80

'''

def Check ():
    global globalstop

    for i in range(10):
        if stri.count(str(i))>repeatcount:
            return False


    if len (stri)==allcount:
        #print("hi",stri)
        Num=str(int(stri)**2)
        if len(Num)==mulcount:
            count=0 
            for i in range(10): 
                if Num.count(str(i))!=totalcount: 
                    break 
                else: 
                    count+=1 

            if count>1:
                print("stri=",stri,count,int(stri)**2)
                print(time.time()-time1)

            if count==10: 
                print("Found",stri)
                print(time.time()-time1)
                globalstop=False

    else:
        temp=stri 
        #Len=len(stri)-1
        
        add=allcount-len(stri) 

        for i in range(add): 
            temp+='9'

        Num1=str(int(temp)**2)

        if len(Num1)!=mulcount:
            print("Electing=",stri)
            return False
        
        
        keep=True
    
        for j in range(9):

            temp3=stri

            for i in range(add):
                temp3+=str(j)

            temp1=int(temp3)**2
            #print(j,temp1)

            if str(temp1)[0]=='1':
                keep=False
                break
            else:
                print("Electing=",stri,j,temp1)

            if keep==False:
                break
            

        count=0
        Num2=str(temp1)

        #print(stri,Num1,Num2,j)

        for i in range(len(Num1)):
            if Num1[i]==Num2[i]:
                count+=1
            else:
                break

        Len=count

    
    

        if Len>0:   
            result=Num1[0:Len] 
    
        else:
            return False


    



        for i in range(10): 
            if result.count(str(i))>totalcount: 
                return False

        #print(stri,len(stri),result)


        return True

def Solve():
    global stri,globalstop

    if globalstop==False:
        return

    for i in range(10):
        stri+=str(i)

        if Check()and globalstop:
            if len(stri)<allcount:
                Solve()
        stri=stri[0:len(stri)-1]

print("Starting with=",stri)
Solve()
print(time.time()-time1)

Actually, some variables should be changed to achieve the solution of the problem. This is kind of clever algorithm and it is obviously a recursive depth search. The crucial part of the algorithm is in the Check function.

What is beautiful here is to get the answer using Python, sure slower than C++ and Fast Fourier version but I have found the smallest square number with a proof.