Prove-that-3-m-3-n-1-is-not-a-perfect-square-where-m-and-n-are-positive-integers-

Question Number 34429 by $@ty@m last updated on 06/May/18

P r o v e t h a t

3^{m} + 3^{n} + 1 i s n o t a p e r f e c t s q u a r e .

w h e r e m a n d n a r e p o s i t i v e i n t e g e r s .

Answered by tanmay.chaudhury50@gmail.com last updated on 06/May/18

l e t a s s u m e m = 2 k

= {(3^{k})}^{2} + 2 . 3^{k} . 1 + 1^{2} + 3^{n} - 2 . 3^{k}

= {(3^{k} + 1)}^{2} + 3^{k} \times (3^{n - k} - 2)

f o r p e r f e c t s q u a r e 3^{k} (3^{n - k} - 2) = 0

s o 3^{n - k} = 2 f o r a n y v a l u e o f n a n d k

t h e t h e v a l u e o f 3^{n - k} c a n n o t b e 2 .

s o 3^{m} + 3^{n} + 1 c s n n o t b e p e r f e c t s q u a e

Commented by $@ty@m last updated on 06/May/18

N i c e a t t e m p t .

b u t i n c o m p l e t e b e c a u s e m = 2 k

m e a n s m i s e v e n .

W h a t i f b o t h m a n d a r e o d d ?

Commented by Rasheed.Sindhi last updated on 07/May/18

{It}^{'} s not necessary to become

3^{k} (3^{n - k} - 2) = 0 for being {(3^{k} + 1)}^{2} + 3^{k} \times (3^{n - k} - 2)

square .

because sum of a square & a number

(although {it}^{'} s not 0) may be a square .

For example

if {(3^{k} + 1)}^{2} = 25 & 3^{k} \times (3^{n - k} - 2) = 11

{(3^{k} + 1)}^{2} + 3^{k} \times (3^{n - k} - 2) = 36

Answered by Rasheed.Sindhi last updated on 08/May/18

3^{m}, 3^{n} & 1 are odd numbers .

∵ The sum of three odd numbers is odd

∴ 3^{m} + 3^{n} + 1 is an odd number .

And if it were perfect square, it must have been

square of an odd number . [Square of an

even number is even of course .]

Let 3^{m} + 3^{n} + 1 = {(2 k + 1)}^{2}

3^{m} + 3^{n} + 1 = {4 k}^{2} + 4 k + 1

3^{m} + 3^{n} = {4 k}^{2} + 4 k

3^{m} + 3^{n} = 4 (k^{2} + k)

k^{2} + k \in E, so right side is divisible by 8

whereas left side is not divisible by 8

{(See proof at the end)}^{*}

∵ 3^{m} + 3^{n} + 1 cannot be square of even nunmber .

because it an odd number .

∵ 3^{m} + 3^{n} + 1 cannot be square of odd nunmber

due to the above contrdiction .

∴ 3^{m} + 3^{n} + 1 {can}^{'} t be perfect square .

- - - - - - - - - - -

* 3^{m} (\mod 8) = 1 or 3

3^{m} + 3^{n} (\mod 8) = 2, 4, 6 only

∴ 3^{m} + 3^{n} (\mod 8) \neq 0

I - e 3^{m} + 3^{n} is not divisible by 8

Commented by MJS last updated on 09/May/18

great!

Commented by MrW3 last updated on 14/Jun/18

w o n d e r f u l p r o o f!

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