Number Magic

Thanks to a post from one of my old manager, mentor and friend - Anand Mohanram - 10 (Square + Cube pattern - Interesting Number Pattern), I was able to find an interesting pattern. Some were commented to his facebook posts.

1. The difference between the cube numbers over their previous square partners is an arithmetic progression with common difference of 1. See them below:

2+1=3

3+3=6

4+6=10

5+10=15

6+15=21

7+21=28

8+28=36

9+36=45

10+45=55

11+55=66

12+66=78

13+78=91

14+91=105

15+105=120

16+120=136

17+136=153

18+153=171

19+171=190

20+190=210

21+210=231

22+231=253

23+253=276

24+276=300

25+300=325

26+325=351

27+351=378

28+378=406

29+406=435

30+435=465

31+465=496

32+496=528

33+528=561

34+561=595

35+595=630

36+630=666

37+666=703

38+703=741

39+741=780

40+780=820

41+820=861

42+861=903

43+903=946

44+946=990

45+990=1035

Actually, we can easily find the numbers matching the pattern by solving the equation:
a^3+b^2 = (a+b)^2;
=>a^3+b^2 = a^2+b^2+2*a*b;
=>b = a*(a-1)/2

This explains the AP of constant 1 between the numbers for every value of a.

Java Source code that helped to find this (though its of no use now) :

public void findSquareCubeNumbers(int count){
 for (int i = 1; i < count; i++){
  int num = i*i*i;
  for (int j = 1; j < count; j++){
   int sum = i+j;
   if (num+(j*j)==sum*sum){
    System.out.println(i+"+"+j+"="+sum);
   }
  }
 }
}

2. For any number n from 1 to 10 and any number k that satisfy the condition - k^n + k^(n+1) = (k+k)^n

the next such condition is satisfied by (2k+1).

The below illustration is the proof:

(k= 1((2*0)+1); n= 1) 1^1+1^2=2^1

(k= 3((2*1)+1); n= 2) 3^2+3^3=6^2

(k= 7((2*3)+1); n= 3) 7^3+7^4=14^3

(k= 15((2*7)+1); n= 4) 15^4+15^5=30^4

(k= 31((2*15)+1); n= 5) 31^5+31^6=62^5

(k= 63((2*31)+1); n= 6) 63^6+63^7=126^6

(k= 127((2*63)+1); n= 7) 127^7+127^8=254^7

(k= 255((2*127)+1); n= 8) 255^8+255^9=510^8

(k= 511((2*255)+1); n= 9) 511^9+511^10=1022^9

(k= 1023((2*511)+1); n= 10) 1023^10+1023^11=2046^10

Java Source code that helped to find this:

public void findNextPowerNumbers(int powVal, int count){
 double prevK = 0;
 for(double i = 1; i < powVal; i++){
  double power = i+1;
  for (double k = 1; k < count; k++){
   double num = Math.pow(k, power);
   double num1 = Math.pow(k, i);
   double sum = 2*k;
   double num3 = Math.pow(sum, i);
   if (num+num1 == num3){
    System.out.println("(k= "+k+"((2*"+prevK+")+1); n= "+i+") "+k+"^"+i+"+"+k+"^"+power+"="+sum+"^"+i);
    prevK=k;
   }
  }
 }
}