show that 3^a + 3^b + 1 is never a perfect square

a, b are natural numbers

Question

show that 3^a + 3^b + 1 is never a perfect square

a, b are natural numbers

anonymous · Answer

pro is writing

parthkohli · Answer

Just throwing out stupid ideas. We want to show this is not possible.$$3^a + 3^b = (k+1)(k-1)$$

anonymous · Answer

If the first term is not a perfect square, the third term is not a perfect square, the middle term is not either 2 or −2 times the product of the square root of the first term and the square root of the third term, then it means it is not a perfect square.

parthkohli · Answer

Please evaluate it before you copy and paste.

rsadhvika · Answer

I guess from that we may sthat $k$ has to be odd

anonymous · Answer

don't look at me i am chust warming up here

imqwerty · Answer

perfect squares end in either 1/4/9/6/5/0

and we have this-
$3^a+3^b+1$$$(2+1)^a +(2+1)^b+1$$$\large\underbrace{\sum_{r=0}^{a} ~^aC_r(2)^r + \sum_{r=0}^{b}~^bC_r(2)^r} +1$ 
  We need to prove that last digit of this part is
                     either 1/2/6/7

that part is even for sure so therefore we just gotta prove that its last digit is either 2 or 6 :/

empty · Answer

either $a=b$ or $a>b$ so those are our two cases to check. Mod 3 arithmetic we have:

$$0^2=0$$$$1^2=1$$$$2^2=1$$

so we know:

$$3^a+3^b+1 = (3n+\alpha)^2$$
where $\alpha = 1$ or $\alpha = 2$. Work it out a little bit and we get:

$$3^a+3^b+1 = 9n^2+6n \alpha + \alpha^2$$

This leads to two cases after we plug in our two choices of alpha,

case 1:
$$3^a+3^b+1 = 9n^2+6n + 1$$$$3^{a-1}+3^{b-1} = n(3n+2)$$
Since $a \ge b$ we must have: $n=3^{b-1}$ $$3^{a-b}+1 =  3n+2 $$ Contradiction and also not true mod 3 anyways. Next case!


case 2:
$$3^a+3^b+1 = 9n^2+12n + 4$$$$3^{a-1}+3^{b-1} =(3n+1)(n+1)$$ Since 3n+1 is a factor then that must mean a>b and $n+1 = 3^{b-1}$ This leads to a contradiction:
$$3^{a-b}+1 = 3n+1 $$$$3^{a-b-1} =n $$ 

Gross but works.

rsadhvika · Answer

|dw:1454934125950:dw|

rsadhvika · Answer

@Empty how $a\ge b$ implies $n = 3^{b-1}$ ?

rsadhvika · Answer

@imqwerty that is a clever trick, but it doesn't work here see http://www.wolframalpha.com/input/?i=Table%5B3%5Ea%2B3%2B1,+%7Ba,+1,+10%7D%5D the expression 3^a + 3^b + 1 evaluates to numbers ending in all digits

ikram002p · Answer

take mod 4 ... only 4n and 4n+1 can  be perfect squares 
 3^a+3^b+1  mod 4 = (-1)^a+(-1)^b+1 mod 4

cases
a even b even
1+1+1 mod 4 can't be perfect square 
a odd b odd
-1-1+1 mod 4 =-1 mod 4 can't be perfect square


a odd b even (or b odd a even)
-1+1+1=1 mod 4  need to check this case :-\

ikram002p · Answer

i'll try mod 12 
only 0,1,4,9 perfect 4 and 0 are even so not considering them.

3^a+3^b+1 mod 3 =1 mod 3
 3^a+3^b+1 mod 4

see we would have same problem with case a odd and b even :-\

rsadhvika · Answer

Very nice ! mod 8 will do that case too i think

ikram002p · Answer

lets find another way...

rsadhvika · Answer

First we observe that $3^a+3^b + 1$ is odd

rsadhvika · Answer

Recall that an odd perfect square is congruent to 1 in mod 8

rsadhvika · Answer

So it is sufficient if we could establish below :  $$3^a+3^b+1\not\equiv 1 \pmod{8}$$

rsadhvika · Answer

http://math.stackexchange.com/questions/268402/prove-that-odd-perfect-square-is-congruent-to-1-modulo-8

ikram002p · Answer

hahaha damn so by contradiction assuming 3^a+3^b+1 = 1 mod 8
then
3^a+3^b| 8 

brilliant !

rsadhvika · Answer

Easy to see that $3^n \equiv 1,3\pmod{8}$

rsadhvika · Answer

None of the combinations $\{1,3\}$ add up to $0 \pmod{8}$. So we're done !

ikram002p · Answer

cool!