IIT JEE , AIEEE Forum - Mathematics , Algebra

md_2674062

prove that n^2+n+1 is not divisble by 5

pramod6990

Can be easily proved using principle of mathematical induction.........

which i donot think is there in the jee syllabus anymore............

iitkgp_bipin

Yes it can solved by mathematical induction.

Another method : if last digit of n²+n+1 is either 0 or 5 then only it is divisible by 5.

When last digit of n is 0, last digit of n²+n+1 is 1.

When last digit of n is 1, last digit of n²+n+1 is 3.

When last digit of n is 2, last digit of n²+n+1 is 7.

When last digit of n is 3, last digit of n²+n+1 is 3.

When last digit of n is 4, last digit of n²+n+1 is 1.

When last digit of n is 5, last digit of n²+n+1 is 1.

When last digit of n is 6, last digit of n²+n+1 is 3.

When last digit of n is 7, last digit of n²+n+1 is 7.

When last digit of n is 8, last digit of n²+n+1 is 3.

When last digit of n is 9, last digit of n²+n+1 is 1.

In any of the cases last digit is neither 0 nor 5, hence n²+n+1 is not divisible by 5.

md_2674062

does any euclid algorithm method

hsbhatt

Method I:

Any number can be written in the form 5q+r where 0

r<5

n²+n+1 = n(n+1)+1 = (5q + r)(5q + r+1)+1

From this we can see that if n²+n+1 is to be divisible by 5, then so must r(r+1)+1 = r²+r+1.

It is easy to check for r =0,1,2,3,4 that the expression is not divisible by 5. Hence it is not divisible by 5 for any n.

Method II:

First eliminate the cases where n = 5k or 5k+1 using the above reasoning or by expanding.

So, assume that n is not of the form 5k or 5k+1

Now, consider n⁴-1 which is divisible by 5 for n prime to 5 by Fermat's theorem.

n⁴-1 = (n-1)(n³+n²+n+1). Since n is not of the form 5k+1, n-1 is not divisible by 5. This means that n³+n²+n+1 is divisible by 5.

Now, if n²+n+1 is divisible by 5, then n³+n²+n+1 is divisible by 5 if n³ is divisible by 5 which means n is divisible by 5 which contradicts our assumption that n is not divisible by 5.

Hence n²+n+1 cannot be divisible by 5 for any n.

So, now you can also answer the question: what is the probability that for a random number n, n³+n²+n+1 is divisible by 5?

Method III:

The result is obviously true for n = 5k.

Suppose n

5k

Let's assume that n²+n+1 is divisible by 5. Then (n²+n+1) (n²-n+1) = n⁴+n²+1 is also divisible by 5.

n⁴+n²+1 = (n⁴-1)+n²+2

(n⁴-1) is divisible by 5 by Fermat's theorem. Hence n²+2 is divisble by 5.

Hence n²+2 = 5k or n² = 5k - 2, which is absurd as any square is of the form 5k, or 5k

1

hsbhatt

md please yaar, just one precious word saying its alright or not.

md_2674062

yaar sum to acha nikala par koi or method nikalo which is of tenth level specialy euclid algortihm

hsbhatt

arre bhai, a=bq+r euclid algorith nahi hey to kya hey?

md_2674062

look at my approach n^2+n+1=5k(if possible)

if k is even then contradiction but for k=odd i am confused

l try to solve on this thinking harsh

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