1) $ a\neq 0$ $ b\neq 0$ then c=3q q odd ,and $ (q - 1)(q + 1) = 2{}^a3{}^{b - 2}$ so b=2 and a=3 c=9 Similarly a=4 b=3 c=21; 2) $ a\neq 0$ $ b = 0$ Then $ 2^a + 9 = c^2^$ and a=4, b=0 , c=5; 3) a=0; $ 3^b+9=c^2$ Then a=0 ,b=3 ,c=6;

Two solutions have been left out in the previous solution. These are $ a = 3, b = 3, c = 15$ and $ a = 5, b = 4, c= 51$.

$ 2^a\cdot3^b+3^2=c^2.\\ i)b=0 \Rightarrow 2^a=(c-3)(c+3)\Rightarrow 2^x=c-3, 2^y=c+3,x<y,x+y=a \Rightarrow 2^x(2^{y-x}-1)=6\Rightarrow\\ x=1,y=3\Rightarrow (a,b,c)=(4,0,5).\\\\ ii)b\geq1\Rightarrow c=3k\Rightarrow b\geq2\\ 2^{a}\cdot3^{b-2}=(k-1)(k+1)\\\\ I).a=0\Rightarrow k-1=3^x,k+1=3^y,x<y,x+y=b-2\Rightarrow 3^x(3^{y-x}-1)=2\Rightarrow x=0,y=1\Rightarrow (a,b,c)=(0,3,6). \\II).a\geq1\Rightarrow k=2p+1\Rightarrow 2^{a-2}\cdot3^{b-2}=p(p+1),(p,p+1)=1\Rightarrow$ $ p=2^{a-2},p+1=3^{b-2}$ or $ p+1=2^{a-2},p=3^{b-2}.$ Then we have to solve the ecuation $ 2^x-3^y=\pm1\\ 2^x-3^y=1$ has solutions $ x=1,y=0, x=2,y=1\Rightarrow (a,b,c)\in\{(3,2,9),(4,3,21)\}$ $ 2^x-3^y=-1$ has solutions $ x=1,y=1, x=3,y=2\Rightarrow (a,b,c)\in\{(3,3,15),(4,5,51)\}$ $ \\S: (a,b,c)\in\{(4,0,5), (0,3,6), (3,2,9), (4,3,21), (3,3,15), (4,5,51)\}$

I got the same answers as those posted in the forum but how do we know that these are all the answers?

The solutions are based on this $ 2^a3^b+6=2^c3^d$, 6 is only one time divisible by 2 and 3, so some of the numbers a,b,c,d are automatic less or equal to 1.

delegat wrote: Solve in non-negative integers the equation $ 2^{a}3^{b} + 9 = c^{2}$ a=b ==>6^a=(c-3)*(c+3) 6^t*6^(t+1)=(c-3)*(c+3) a=2t+1 ==> (a,b,c)=(3,3,15);(a,b,c) =(5,5,39);....;(a,b,c)=(2t+1,2t+1,c^((2t+1)/2)-3) ............... highlander wrote: $ 2^a\cdot3^b+3^2=c^2.\\ i)b=0 \Rightarrow 2^a=(c-3)(c+3)\Rightarrow 2^x=c-3, 2^y=c+3,x<y,x+y=a \Rightarrow 2^x(2^{y-x}-1)=6\Rightarrow\\ x=1,y=3\Rightarrow (a,b,c)=(4,0,5).\\\\ ii)b\geq1\Rightarrow c=3k\Rightarrow b\geq2\\ 2^{a}\cdot3^{b-2}=(k-1)(k+1)\\\\ I).a=0\Rightarrow k-1=3^x,k+1=3^y,x<y,x+y=b-2\Rightarrow 3^x(3^{y-x}-1)=2\Rightarrow x=0,y=1\Rightarrow (a,b,c)=(0,3,6). \\II).a\geq1\Rightarrow k=2p+1\Rightarrow 2^{a-2}\cdot3^{b-2}=p(p+1),(p,p+1)=1\Rightarrow$ $ p=2^{a-2},p+1=3^{b-2}$ or $ p+1=2^{a-2},p=3^{b-2}.$ Then we have to solve the ecuation $ 2^x-3^y=\pm1\\ 2^x-3^y=1$ has solutions $ x=1,y=0, x=2,y=1\Rightarrow (a,b,c)\in\{(3,2,9),(4,3,21)\}$ $ 2^x-3^y=-1$ has solutions $ x=1,y=1, x=3,y=2\Rightarrow (a,b,c)\in\{(3,3,15),(4,5,51)\}$ $ \\S: (a,b,c)\in\{(4,0,5), (0,3,6), (3,2,9), (4,3,21), (3,3,15), (4,5,51)\}$

Marius Mainea wrote: 1) $ a\neq 0$ $ b\neq 0$ then c=3q q odd ,and $ (q - 1)(q + 1) = 2{}^a3{}^{b - 2}$ so b=2 and a=3 c=9 Similarly a=4 b=3 c=21; 2) $ a\neq 0$ $ b = 0$ Then $ 2^a + 9 = c^2^$ and a=4, b=0 , c=5; 3) a=0; $ 3^b+9=c^2$ Then a=0 ,b=3 ,c=6; could u explain me the first step in detail manner

mavropnevma wrote: Two solutions have been left out in the previous solution. These are $ a = 3, b = 3, c = 15$ and $ a = 5, b = 4, c= 51$. how did u get to this answer

siddharthanand wrote: Marius Mainea wrote: 1) $ a\neq 0$ $ b\neq 0$ then c=3q q odd ,and $ (q - 1)(q + 1) = 2{}^a3{}^{b - 2}$ so b=2 and a=3 c=9 Similarly a=4 b=3 c=21; 2) $ a\neq 0$ $ b = 0$ Then $ 2^a + 9 = c^2^$ and a=4, b=0 , c=5; 3) a=0; $ 3^b+9=c^2$ Then a=0 ,b=3 ,c=6; could u explain me the first step in detail manner Casework: either $2||q-1 \wedge 2^{a-1}||q+1$ or vice versa, and again $3^{b-2}||q-1$ or $3^{b-2}||q+1$. Mihailescu's theorem makes it easier in some cases. Otherwise $0 \in \{a,b \}$.

highlander wrote: Then we have to solve the ecuation $ 2^x-3^y=\pm1\\ 2^x-3^y=1$ has solutions $ x=1,y=0, x=2,y=1\Rightarrow (a,b,c)\in\{(3,2,9),(4,3,21)\}$ $ 2^x-3^y=-1$ has solutions $ x=1,y=1, x=3,y=2\Rightarrow (a,b,c)\in\{(3,3,15),(4,5,51)\}$ So highlander is using Mihailescu's theorem here to find the only solutions such that x,y > 1 ?

cheeseyicecream wrote: highlander wrote: Then we have to solve the ecuation $ 2^x-3^y=\pm1\\ 2^x-3^y=1$ has solutions $ x=1,y=0, x=2,y=1\Rightarrow (a,b,c)\in\{(3,2,9),(4,3,21)\}$ $ 2^x-3^y=-1$ has solutions $ x=1,y=1, x=3,y=2\Rightarrow (a,b,c)\in\{(3,3,15),(4,5,51)\}$ So highlander is using Mihailescu's theorem here to find the only solutions such that x,y > 1 ? what's Mihailescu's theorem ? I've got exactly the same answer as yours,but the way I achieved it is very complex.

Preda Mihailescu proved Catalan's conjecture that $8$ and $9$ are the only consecutive powers of positive integers.

My solution is a bit long but i didnt miss any cases so here it is: first look for the cases where a,b are 0; a is 0 and b isnt 0, b is 0 and a isnt 0.These should be easy by factorising. And for the case where a,b,c dont equal 0 just divide both sides by 4.You will be left with (c-3)/2 and (c+3)/2.Note that both of them cant be even but they can be a power of 3 at the same time.Lets say a>=3 and b=x+y. The cases we have now are: 1)(c-3)/2=2^(a-2) and (c+3)/2=3^y 2)(c-3)/2=3^x and (c+3)/2=3^(y)2^(a-2) 3)(c-3)/2=1 and (c+3)/2=3^(b)2^(a-2) Then just check the case where a=2 and youre done.

highlander wrote: $ 2^a\cdot3^b+3^2=c^2.\\ i)b=0 \Rightarrow 2^a=(c-3)(c+3)\Rightarrow 2^x=c-3, 2^y=c+3,x<y,x+y=a \Rightarrow 2^x(2^{y-x}-1)=6\Rightarrow\\ x=1,y=3\Rightarrow (a,b,c)=(4,0,5).\\\\ ii)b\geq1\Rightarrow c=3k\Rightarrow b\geq2\\ 2^{a}\cdot3^{b-2}=(k-1)(k+1)\\\\ I).a=0\Rightarrow k-1=3^x,k+1=3^y,x<y,x+y=b-2\Rightarrow 3^x(3^{y-x}-1)=2\Rightarrow x=0,y=1\Rightarrow (a,b,c)=(0,3,6). \\II).a\geq1\Rightarrow k=2p+1\Rightarrow 2^{a-2}\cdot3^{b-2}=p(p+1),(p,p+1)=1\Rightarrow$ $ p=2^{a-2},p+1=3^{b-2}$ or $ p+1=2^{a-2},p=3^{b-2}.$ Then we have to solve the ecuation $ 2^x-3^y=\pm1\\ 2^x-3^y=1$ has solutions $ x=1,y=0, x=2,y=1\Rightarrow (a,b,c)\in\{(3,2,9),(4,3,21)\}$ $ 2^x-3^y=-1$ has solutions $ x=1,y=1, x=3,y=2\Rightarrow (a,b,c)\in\{(3,3,15),(4,5,51)\}$ $ \\S: (a,b,c)\in\{(4,0,5), (0,3,6), (3,2,9), (4,3,21), (3,3,15), (4,5,51)\}$ Umm, you ordered a and b wrong in the last sol… Cuz I got the last pair was (5, 4, 51), which is correct, not (4, 5, 51) for (a, b, c). Also, I got the obvious solutions which were these, but I wasn’t sure if it was the full solution, because $2^x-3^y=\pm1$ I only got the obvious sols of x=1 y=0 and x=2 y=1, but how do we know there are no larger solutions to x and y? (And analogously for the other cases, because I got the same answer as you but didn’t know if it was all solutions) Edit: After reading other solutions I see why now. Glad that at least I got the right answers! The results are from Catalan’s conjecture https://en.wikipedia.org/wiki/Catalan%27s_conjecture

huashiliao2020 wrote: After reading other solutions I see why now. Glad that at least I got the right answers! The results are from Catalan’s conjecture To solve the equation $2^x-3^y= \pm 1$ , Catalan's conjecture is not necessary at all. Try mod 3 and 4