See here.

INMO 1992

Hello everyone! Thats my solution I hope that , it isnt wrong. $2^x+3^y=a²$ Step 1 $x=0 $ $a²-1=3^y$ $(a-1)*(a+1)=3^y$ We know that a is $ODD$ number And by that the only solution for that $a=2;y=1;x=0$ Step 2 $y=0$ within that $a²-1=2^x$ $(a-1)*(a+1)=2^x$ $x=c+l $ $a-1=2^c$ and $ a+1=2^l$ By that $2a=2^c(1+2^(l-c))$ And $$c=1$$By that $$x=3;y=0;a=3$$The Third and last one: By the mod 3 and mod 4 and mod 5 If we do replacement by respectively k,t,u $4^k+9^t=25^d$ $5^(2f)-2^(2z)=9^t$ By the last equation only solution which we get by case is f=1 z=1 t=1 and $$(4,2,5)$$And we do....

Let $2^x + 3^y = z^2$ with $z \in \mathbb{Z+}$. It's not difficult to see that $x \equiv 0 \pmod{2}$ by using modulo $3$. For $x=2$, We have $2^{2} + 3^y = z^2 \rightarrow 3^y = (z-2)(z+2)$. Let $z-2= 3^b$ and $z+2=3^c$ with $b<c$ and $b+c=y$. $3^c - 3^b = (z+2)-(z-2) = 4$. There is no solution for this equation. For $x>2$, by modulo $8$ we have $3^y \equiv 1 \pmod {8}$. Therefore $y$ is even, let $y=2d$ with $d \in \mathbb{Z+}$. $2^x+3^{2d}=z^2 \rightarrow 2^x = (z-3^d)(z+3^d)$. Let $z-3^d=2^e$ and $z+3^d=2^f$ with $ e<f$ and $e+f=x$. $2^f-2^e=(z+3^d)-(z-3^d)=2\cdot3^d$ $\rightarrow 2^{f-1}-2^{e-1}=3^d$. Observe that $RHS$ is an odd number. Because $e<f$, $e$ must be less than 2. Therefore, $e=1$. $\rightarrow 2^{f-1}-1=3^d \rightarrow 2^{f-1} - 3^d = 1$. For $f >4$, by modulo $8$ $\rightarrow 3^d \equiv 7 \pmod{8}$. No such $d$ exist. Therefore, $f\le3$. For $f=3$, $d=1. \rightarrow (x,y,z)=(4,2,5)$. For $f=2$ and $1$, no solution exist. Hence, the only solution for $(x,y)$ is $(4,2)$.

$x$ is even (modulo 3) $y$ is even (modulo 4) $x=2a,y=2b$ and $(2^a,3^b,z)$ is a primitive Pythagorean triple. $\gcd(u,v)=1$ and $2^a=2uv\Rightarrow v=1$ $3^b=u^2-v^2\Rightarrow u-v=1$ $u=2,v=1\Rightarrow (x,y,z)=(4,2,5)$

y=0 için tek çözüm 2^x=(z+1)(z-1) denkleminden (3,0,3) olur. y=>1 için mod 3 te incelersek x=2k ve buradan 3^y=(z+2^k)(z-2^k) olur. z+2^k=3^a dersek 3^a-3^(y-a)=2^(k+1) olur.y=a olmak zorunda olduğundan 3^a-1=2^(k+1) olur.Catalan teoreminden dolayı gelen (a,k) çözümleri (1,0) ve (2,2);(x,y,z) çözümleri (0,1,2) ve (4,2,2) gelir.

Davut1102 wrote: y=0 için tek çözüm 2^x=(z+1)(z-1) denkleminden (3,0,3) olur. y=>1 için mod 3 te incelersek x=2k ve buradan 3^y=(z+2^k)(z-2^k) olur. z+2^k=3^a dersek 3^a-3^(y-a)=2^(k+1) olur.y=a olmak zorunda olduğundan 3^a-1=2^(k+1) olur.Catalan teoreminden dolayı gelen (a,k) çözümleri (1,0) ve (2,2);(x,y,z) çözümleri (0,1,2) ve (4,2,2) gelir. Pozitif tamsayı istiyorsa (0,1,2) çözüm değildir

A square of integer is equal to 0,1 for mod 4. So we can search it for cases; Search Pow(2,x); There are two cases such as x=1 and x>1 i) x=1 ----> 2==2(mod 4) And also pow(3,y)==1,3 (mod 4) ii) x>=2 pow(2,x)==0(mod 4) From there we can comprehend that we must use y which is even. We get a answer (x,y)---->(4,2)

Can somebody please tell me if this is a fakesolve? oops my woot writing problem had variables m,n instead of x,y (just assume that the m,n are x,y) We claim that $(m,n)=\boxed{(4,2)}$ is the only solution that works. Note that all residues of squares $\pmod 3$ are $0,1 \pmod 3.$ Now note that if $m=1,$ we have $2+3^n \equiv 2 \pmod 3,$ which can't be a square since all residues are $0,1 \pmod 3.$ Claim: If $m>1,$ then $n, m$ has to be even. Proof: If $n$ is odd, note that taking $\pmod 4,$ we have $2^m +3^n \equiv 3^n\equiv 3,1 \pmod 4,$ if $n$ is odd then $3^n \equiv 3 \pmod 4.$ Note that all squares $\pmod 4$ are $0,1 \pmod 4,$ thus we have a contradiction and $n$ is even. Now note that if $m$ is odd, then $2^m \equiv 2 \pmod 3,$ so $2^m+3^n \equiv 2 \pmod 3,$ but all perfect squares $\pmod 3$ are $0,1 \pmod 3$ resulting in a contradiction. Since both $m,$ and $n$ are even, then $2^m$ and $3^n$ are perfect squares, thus letting $m=2a,$ and $n=2b,$ thus we want $(2^a)^2+(3^b)^2=c^2,$ for some integer $c.$ Thus we want $(c-2^a)(c+2^a)=3^{2b}.$ We can take casework on $c-2^a,$ and $c+2^a.$ Case 1: If $c-2^a, c+2^a>1.$ Then $c-2^a, c+2^a \equiv 0\pmod 3,$ thus we have $(c+2^a)-(c-2^a)= 2^{a+1} \equiv 0\pmod 3,$ but this can't be true as a power of $2$ has no factor of $3,$ thus we have a contradiction. Case 2: If one or both of $c-2^a, c+2^a=1.$ In this case, note that $c+2^a>c-2^a$ so this means that $c-2^a=1,$ and $c+2^a=3^{2b}.$ So subtracting gives $3^{2b}-1=2^{a+1},$ so $(3^b-1)(3^b+1)=2^{a+1},$ thus we can deduce that $x=(3^b-1)$ and $y=(3^b+1)$ are both powers of $2.$ Now since $x,y$ are two powers of two that differ by $2,$ we claim that $x=2,$ and $y=4$ is the only solution that works. Now let $x=2^w,$ and $y=2^z.$ Without loss of generality we assume $y>x.$ Now since this difference is even, either both $x,y$ are odd which results in a contradiction, or both are even. Since both are even, then we have $2^{z-1}-2^{w-1}=1.$ Thus one of the powers is odd. So we must have either $z,w$ be $1.$ If $z=1$ then $2^{w-1}=0$ resulting in a contradiciton. If $w=1$ then $2^{z-1}=2$ thus $z-1=1$ and $z=2.$ So we have that $x=2,$ and $y=4$ is the only solution that works. Thus $2=3^b-1$ and $4=3^b+1$ which implies $b=1.$ So $n=2$ and we know that $(c-2^a)(c+2^a)=9.$ Now we can deduce $c+2^a=9$ and $c-2^a=1,$ subtracting yeilds $2^{a+1}=8,$ so $2^a=4$ and $a=2.$ So $m=2a=4.$ Thus $(m,n)=\boxed{(4,2)}$ is the only solution that works.

parmenides51 wrote: Find all pairs of positive integers $(x,y)$ such that $2^x + 3^y$ is a perfect square. With $mod3$ we get that $x=0(mod2)$ With $mod4$ we get that $y=0(mod2)$ So let $x=2a$ and $y=2b$ We want to solve the equation $2^{2a}+3^{2b}=z^2$ or $3^{2b}=(z-2^a)(z+2^a)$ We can se now that $(z-2^a,z+2^a)=1$ so we are going to have: $z-2^a=1$ and $z+2^a=3^{2b}$ This becames $2^{a+1}+1=3^{2b}$ or $2^{a+1}=(3^b-1)(3^b+1)$ Now we can see that $(3^b-1,3^b+1)=2$ so we have that: $3^b-1=2$ and $3^b+1=2^a$ From here esily we get the only sollution $(x,y)=(4,2)$