IIT JEE , AIEEE Forum - Mathematics , Algebra

md_2674062

. Let n be an integer. Show that, if 2 + 2sqrt(1 + 12n^2)is an integer, then

it is a perfect square

nadeemoidu

If the given expression is to be an integer , then (1 + 12n²) should be a perfect square.

let (1 + 12n²) = ( 2k + 1 ) ² [ Its obvious that the no. should be odd]

12n² = 4k(k+1)

3n² = k(k+1)

now , k and k+1 are coprime.

so either k should be a multiple of n² or k+1 should be a multiple of n²
......

Can someone complete the answer ? I can't.

md_2674062

thanks yaar but i can understand ur last steps ie the value of k or k+1 how r u taking that is there is ant other way

md_2674062

for n=28 u will get another ans .so n=2 is only one option

siddhartha100

As answered by nadeemoidu

3n²=k(k+1)

3=k(k+1)/n²...........................................1

As k & k+1 have no common factors, either k or k+1 is divisible by n²

let k=xn² for some integer x

so (1) comes to 3=x(xn²+1)

So either x=3 & xn²+1=1 OR x=1 & xn²+1=3 [as 3 is prime]

in both the cases n is not an integer

Hence k+1=xn²

substituting k+1 in (1) 3=x(xn²-1)

for x=1 and xn²-1=3 n²=4, which is the only soln.

So 2+2sqrt(1+12n²)=16=4²

hsbhatt

nadeemoidu

@siddartha100
I had typed an almost similar answer completely but after it was pointed out that n= 28 is also an option , I deleted it. there is some fault in our reasoning.

If we check the case when n=28, we can see that it is not necessary that k or k+1 is divisible by n. It is possible that n= pq where p and q are 2 integers and
k=3p² and k+1 = q² ,etc .

Now I don't think that we can exhaustively find all the possible values for n , we will have to find a theoratical answer.

hsbhatt

see this link:
http://www.mathlinks.ro/viewtopic.php?t=150375

hsbhatt

We must have 1+12n² = p² where p is odd.

p² - 1 = 12n² or (p-1)(p+1) = 12n^2.

If p = 2k+1 then k(k+1) = 3n^2_.

Now, either k or k+1 must be divisible by 3.

Lets say k+1 is div by 3. Then k = 3m-1 where m

Z.

Then m(3m-1) = n². Now gcd(m,3m-1) = 1

Hence, m and 3m-1 are themselves perfect squares. But we know that any perfect square is of the form 3t or 3t+1.

Hence 3m-1 cannot be a perfect square and our assumption that k+1 is div by 3 is wrong.

This means k is div by 3 i.e k = 3m

Then m(3m+1) = n²

and again since gcd(m,3m+1) = 1, we must have m = s₁²and 3m+1 = s₂²

Hence, sqrt(1+12n²) = p = 2k+1 = 6m+1 = 6s₁²+1

Hence 2(1+p) = 2(6s₁²+2) = 4(3s₁²+1) = 4(3m+1) = 4s₂² = a perfect square.

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