Summation vs Square 2

Number Theory Level 4

∑ k = 1 N k 2 = m 2 \displaystyle \sum_{k=1}^Nk^2=m^2

Find the least value of N > 1 N>1 .Where m ϵ Z \quad m\epsilon Z


The answer is 24.

Kalpok Guha by Kalpok Guha , 21, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

S e t i n c a l c u l a t o r t h e f o l l o w i n g . ∑ 1 n x 2 . F e e d n = 2 , 3 , 4...... t i l l t h e a n s w e r i s a n i n t e g e r . F o r n = 24 , t h e a n s w e r = 70. T h i s m e a n s s u m 1 + 2 + 3... + 69 + 70 = 24 2 . Set~in~calculator~the ~following.\\ \displaystyle \sqrt{\sum_1^n x^2}.\\ Feed~n=2,3,4......~~till ~the~answer~is~an~integer.\\ For~n=24,~the~answer=70.\\ This~means~sum~~ 1+2+3 . . .+69+70=\huge \color{#D61F06}{24}^2. \\ ~~~~\\ ~~~~~\\

A program could be:- n = 1 ∗ ∗ n = n + + x = n ∗ ( n + 1 ) ∗ ( 2 n + 1 ) 6 I f x − ⌊ x ⌋ = 0 p r i n t n r e t u r n E l s e g o ∗ ∗ n=1\\ **n=n++\\ x= {\sqrt{\dfrac{n*(n+1)*(2n+1)} 6}}\\ If~~ x-\lfloor x \rfloor=0\\ ~~~~~~print~n\\ ~~~~~~return\\ Else~go~**

0 Helpful 0 Interesting 0 Brilliant 0 Confused
David Holcer
Mar 12, 2015 
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
 
from math import *
found=False
num=2

while not found:
    a=num**3/float(3)+num**2/float(2)+num/float(6)
    if sqrt(a).is_integer():
        print "Found %s"%num
        found=True
    num+=1

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...