Programming a number theory programme

Question

I have this thing that I want to programme which goes as follows.

We all know the next two identities:

3 ² +4 ² =5 ²
3 ³ +4 ³ +5 ³ =6 ³

Now, I want to write a computer code (in mathematica,C or Python) that will check for such relations.

Eg, for 3 ⁴ +4 ⁴ +5 ⁴ +6 ⁴ compare it with 7 ⁴ and check if it equals to it or not, I mean I want to check for more cases of such a sequence of numbers as above and compare them if indeed they make a sequence as above.

Basically I know I need here a loop and conditionals, my problem is how do I keep the numbers 3,4,5,6,... to keep being generated in the sequences?

This is where I am bugged down as to how to write this code.

I mean I would like to check for upto i=10,000, ie: 3 ⁱ +4 ⁱ +5 ⁱ +... does it equal (3+i) ⁱ etc...

I hope you understood my question.

Thanks in advance.

Answer 1

for pow in xrange(2,5):
    sum=0
    for index in xrange(3,3+pow):
        sum+=index ** pow
    if sum==(index+1) ** pow:
        print True

#output is for power 2, 3 it works

i took it for range of 2,5

ie first loop calculates 3**2 + 4 **2 ==5 **2 ..soo on

increase the pow range to 10001 for all 10000 powers

Answer 2

mean I would like to check for upto i=10,000, ie: 3i+4i+5i+... does it equal (3+i)i etc...

Accounting for the way range works, and printing only when True :

limit = 10000
for pow in range(2, limit + 1):
    if ((3 + pow) ** pow == sum([exp ** pow for exp in range(3, 3 + pow)])):
        print pow

Answer 3

This tries to avoid re-doing calculations as much as possible.

>>> def f(n):
...    c = 1
...    L1 = [3]
...    L2 = [1]
...    while (c + 3 < n):
...       L2 = [L1[i] * L2[i] for i in range(c)]
...       c += 1
...       x = (c + 2) ** (c - 1)
...       print(c, x == sum(L2))
...       L1.append(c + 2)
...       L2.append(x)
...
>>> f(10)
(2, False)
(3, True)
(4, True)
(5, False)
(6, False)
(7, False)
>>>

Answer 4

Well, in Mathematica it's quite easy:

For[i = 3, i < 101, i++ ]
If[Sum[(3 + j)^i, {j, 0, i - 1}] == (3 + i)^i, Print[i]]

I am pretty much sure that in other programming languages it may run faster, though.