Solve in nonnegative integers the equation $5^t + 3^x4^y = z^2$.
Problem
Source: JBMO Shortlist 2017 NT4
Tags: number theory, Diophantine equation
TuZo
25.07.2018 17:39
Hint: If $t=2u+1$, we have LHS is $3k+2$, and this is contradiction with RHS. (Excepting $x=0$) So $t=2u$, thus we have $z-{{5}^{u}}={{3}^{a}}\cdot {{4}^{b}},z+{{5}^{u}}={{3}^{c}}\cdot {{4}^{d}}$, etc.
Sevil1
13.03.2020 13:07
Is this problem solution only {t=0 x=1 y=1 z=2}?
Sevil1
13.03.2020 13:28
please can someone help me ?
Raunii
24.03.2020 09:15
Sevil1 wrote: Is this problem solution only {t=0 x=1 y=1 z=2}? I think there are no solutions and the solution you mentioned is not satisfying the equation
ematruc
24.03.2020 13:19
I found 3 solutions $(t,x,y,z)$:
$(1,0,1,3), \ (0,1,0,2) \ and \ (0,1,2,7)$
The proof I found is bashy and long but here it is:
As TuZo suggested $t$ must be even except for $x=0$.
So if $x=0$ we have $$5^t+(2^2)^y=z^2 $$$$ 5^t=(z-2^y)(z+2^y)$$Since the GCD of the factors in RHS is $1$ (because we get that it's the GCD between $(z+2^y)$ that is odd and $2^{y+1}$), we have that one of the factors is $1$ so $z=2^y+1$ and substituting we get $$5^t=2^{y+1}+1$$that by Michailescu has solutions iff $t$ or $y+1$ is one and so only the case $5+4=3^2$ that is the first solution.
Now we can substitute $t=2v$ in the first equation and factor to get $$2^{2y}3^x=(z-5^v)(z+5^v)$$. If $y=0$ we have $$3^x=(z-5^v)(z+5^v)$$so $z-5^v=1$ and $3^x=2 \cdot 5^v +1$ and we have the second solution $x=1 \ \wedge \ v=0$. If $v>0$ the last digit of $3^x$ is $1$ and we can write $x=4n$. So we get $$81^n-1=2 \cdot 5$$That is impossible because $80 \nmid 2 \cdot 5^v$
Again find the GCD between the RHS factors and you get it's $2$ (because it's $GCD[(z+5^v);(2 \cdot 5^v)]$ and z is odd if $y>0$ and not multiple of $5$) and so you have to divide the factors $2, 2^{2y-1}, 3^x$ between the two factors. (The factors $2$ and $2^{2y-1}$ are split between the 2 factors and $3^x$ could stay in both factors so we have 4 cases.) Three of these cases won't lead anywhere and the fourth that is $$\begin{cases}
z+5^v=2^{2y-1} \\
z-5^v=2 \cdot 3^x
\end{cases}$$will give the third solution. (All these proofs are similar to the previous ones so I will omit them.)
Telman
31.05.2022 22:19
ematruc wrote: I found 3 solutions $(t,x,y,z)$:
$(1,0,1,3), \ (0,1,0,2) \ and \ (0,1,2,7)$
The proof I found is bashy and long but here it is:
As TuZo suggested $t$ must be even except for $x=0$.
So if $x=0$ we have $$5^t+(2^2)^y=z^2 $$$$ 5^t=(z-2^y)(z+2^y)$$Since the GCD of the factors in RHS is $1$ (because we get that it's the GCD between $(z+2^y)$ that is odd and $2^{y+1}$), we have that one of the factors is $1$ so $z=2^y+1$ and substituting we get $$5^t=2^{y+1}+1$$that by Michailescu has solutions iff $t$ or $y+1$ is one and so only the case $5+4=3^2$ that is the first solution.
Now we can substitute $t=2v$ in the first equation and factor to get $$2^{2y}3^x=(z-5^v)(z+5^v)$$. If $y=0$ we have $$3^x=(z-5^v)(z+5^v)$$so $z-5^v=1$ and $3^x=2 \cdot 5^v +1$ and we have the second solution $x=1 \ \wedge \ v=0$. If $v>0$ the last digit of $3^x$ is $1$ and we can write $x=4n$. So we get $$81^n-1=2 \cdot 5$$That is impossible because $80 \nmid 2 \cdot 5^v$
Again find the GCD between the RHS factors and you get it's $2$ (because it's $GCD[(z+5^v);(2 \cdot 5^v)]$ and z is odd if $y>0$ and not multiple of $5$) and so you have to divide the factors $2, 2^{2y-1}, 3^x$ between the two factors. (The factors $2$ and $2^{2y-1}$ are split between the 2 factors and $3^x$ could stay in both factors so we have 4 cases.) Three of these cases won't lead anywhere and the fourth that is $$\begin{cases}
z+5^v=2^{2y-1} \\
z-5^v=2 \cdot 3^x
\end{cases}$$will give the third solution. (All these proofs are similar to the previous ones so I will omit them.)
Very good solution thanks for posting but i think that u missed a solution: (t,x,y,z)=(2,2,2,13)
amogususususus
06.12.2022 12:33
can anyone tell me what is Mihailescu theorem, what i know is that Mihailescu theorem is the same as catalan conjecture that states: $8=2^3$ and $9=3^2$ are the only consecutive power integer.
Davut1102
06.12.2023 21:48
Denklem 5^t+(3^x)(4^y)=z^2 şeklinde. z^2=0, 1(mod 4) olduğundan dolayı t=2d formundadır. Bu durumda (3^x) (4^y) =(z+5^d) (z-5^d) olur. z+5^d=(3^a) (2^b) denildiğinde z-5^d=[3^(x-a)][2^(2y-b)] olur ve buradan 2.5^d=(3^a)(2^b)-[3^(x-a)][2^(2y-b)] denklemini çözmemiz gerekir. mod 3 te incelersek x=a olması gerekir ve denklem 2.5^d=(3^a)(2^b)-2^(2y-b) halini alır. b=0 ise 2.5^d=3^a-2^2y olur ve mod 2 de incelendiğinde y=0 olur ve denklem 2.5^d=3^a-1 halini alır. İfade sırasıyla 4 ve 5 modlarında incelenirse a nın hem tek hem çift olması gerekir, çelişki. b=1 ise yerine yazıp iki tarafı da ikiye bölersek 5^d=3^a-2^(2y-1) denklemine ulaşırız.