Mixing
Solutions of $2^a+5^b=c^2$
$begingroup$
I tried to solve this question using this way:
$a$ must be even, because if it's odd the equation have no solution. Let $a=2n$, so
begin{align*}
2^{2n}+5^b &= c^2 \
(2^n)^2+5^b &= c^2 \
5^b &= c^2-(2^n)^2 \
5^b &= (c-2^n)(c+2^n)
end{align*}
Only one of $c-2^n$ and $c+2^n$ can be divided by 5, $c-2^n neq c+2^n$, so $c-2^n=1$ and $c+2^n=5^b$.
From this equations, I got $1+2^{n+1}=5^b$.
But I don't know how to continue.
number-theory
edited Jan 2 at 4:39
Robert Israel
319k23209459
asked Jan 2 at 4:25
Agung Izzul HaqAgung Izzul Haq
563
$endgroup$
|
show 4 more comments
$begingroup$
I tried to solve this question using this way:
$a$ must be even, because if it's odd the equation have no solution. Let $a=2n$, so
begin{align*}
2^{2n}+5^b &= c^2 \
(2^n)^2+5^b &= c^2 \
5^b &= c^2-(2^n)^2 \
5^b &= (c-2^n)(c+2^n)
end{align*}
Only one of $c-2^n$ and $c+2^n$ can be divided by 5, $c-2^n neq c+2^n$, so $c-2^n=1$ and $c+2^n=5^b$.
From this equations, I got $1+2^{n+1}=5^b$.
But I don't know how to continue.
number-theory
edited Jan 2 at 4:39
Robert Israel
319k23209459
asked Jan 2 at 4:25
Agung Izzul HaqAgung Izzul Haq
563
$endgroup$
$begingroup$
@Randall: If $a$ is odd, then the solution fails mod $4$.
$endgroup$
– JavaMan
Jan 2 at 4:32
1
$begingroup$
a, b, and c are integers.If $a$ is odd, then the last digit of $2^a$ is 2 or 8. Added by 5, we got 7 or 3. There is no integer $c$ such that the last digit of $c^2$ is 7 or 3. I forgot to state that the solutions must be integer.
$endgroup$
– Agung Izzul Haq
Jan 2 at 4:32
$begingroup$
Got it. My fault.
$endgroup$
– Randall
Jan 2 at 4:33
$begingroup$
@Randall: Took me a second to see it too!
$endgroup$
– JavaMan
Jan 2 at 4:33
6
$begingroup$
Well, you could use Mihăilescu's theorem.
$endgroup$
– Robert Israel
Jan 2 at 4:38
|
show 4 more comments
$begingroup$
I tried to solve this question using this way:
$a$ must be even, because if it's odd the equation have no solution. Let $a=2n$, so
begin{align*}
2^{2n}+5^b &= c^2 \
(2^n)^2+5^b &= c^2 \
5^b &= c^2-(2^n)^2 \
5^b &= (c-2^n)(c+2^n)
end{align*}
Only one of $c-2^n$ and $c+2^n$ can be divided by 5, $c-2^n neq c+2^n$, so $c-2^n=1$ and $c+2^n=5^b$.
From this equations, I got $1+2^{n+1}=5^b$.
But I don't know how to continue.
number-theory
edited Jan 2 at 4:39
Robert Israel
319k23209459
asked Jan 2 at 4:25
Agung Izzul HaqAgung Izzul Haq
563
$endgroup$
I tried to solve this question using this way:
$a$ must be even, because if it's odd the equation have no solution. Let $a=2n$, so
begin{align*}
2^{2n}+5^b &= c^2 \
(2^n)^2+5^b &= c^2 \
5^b &= c^2-(2^n)^2 \
5^b &= (c-2^n)(c+2^n)
end{align*}
Only one of $c-2^n$ and $c+2^n$ can be divided by 5, $c-2^n neq c+2^n$, so $c-2^n=1$ and $c+2^n=5^b$.
From this equations, I got $1+2^{n+1}=5^b$.
But I don't know how to continue.
number-theory
number-theory
edited Jan 2 at 4:39
Robert Israel
319k23209459
asked Jan 2 at 4:25
Agung Izzul HaqAgung Izzul Haq
563
edited Jan 2 at 4:39
Robert Israel
319k23209459
asked Jan 2 at 4:25
Agung Izzul HaqAgung Izzul Haq
563
edited Jan 2 at 4:39
Robert Israel
319k23209459
edited Jan 2 at 4:39
Robert Israel
319k23209459
edited Jan 2 at 4:39
Robert Israel
319k23209459
319k23209459
asked Jan 2 at 4:25
Agung Izzul HaqAgung Izzul Haq
563
asked Jan 2 at 4:25
Agung Izzul HaqAgung Izzul Haq
563
asked Jan 2 at 4:25
Agung Izzul HaqAgung Izzul Haq
563
563
$begingroup$
@Randall: If $a$ is odd, then the solution fails mod $4$.
$endgroup$
– JavaMan
Jan 2 at 4:32
1
$begingroup$
a, b, and c are integers.If $a$ is odd, then the last digit of $2^a$ is 2 or 8. Added by 5, we got 7 or 3. There is no integer $c$ such that the last digit of $c^2$ is 7 or 3. I forgot to state that the solutions must be integer.
$endgroup$
– Agung Izzul Haq
Jan 2 at 4:32
$begingroup$
Got it. My fault.
$endgroup$
– Randall
Jan 2 at 4:33
$begingroup$
@Randall: Took me a second to see it too!
$endgroup$
– JavaMan
Jan 2 at 4:33
6
$begingroup$
Well, you could use Mihăilescu's theorem.
$endgroup$
– Robert Israel
Jan 2 at 4:38
|
show 4 more comments
$begingroup$
@Randall: If $a$ is odd, then the solution fails mod $4$.
$endgroup$
– JavaMan
Jan 2 at 4:32
1
$begingroup$
a, b, and c are integers.If $a$ is odd, then the last digit of $2^a$ is 2 or 8. Added by 5, we got 7 or 3. There is no integer $c$ such that the last digit of $c^2$ is 7 or 3. I forgot to state that the solutions must be integer.
$endgroup$
– Agung Izzul Haq
Jan 2 at 4:32
$begingroup$
Got it. My fault.
$endgroup$
– Randall
Jan 2 at 4:33
$begingroup$
@Randall: Took me a second to see it too!
$endgroup$
– JavaMan
Jan 2 at 4:33
6
$begingroup$
Well, you could use Mihăilescu's theorem.
$endgroup$
– Robert Israel
Jan 2 at 4:38
$begingroup$
@Randall: If $a$ is odd, then the solution fails mod $4$.
$endgroup$
– JavaMan
Jan 2 at 4:32
$begingroup$
@Randall: If $a$ is odd, then the solution fails mod $4$.
$endgroup$
– JavaMan
Jan 2 at 4:32
1
1
$begingroup$
a, b, and c are integers.If $a$ is odd, then the last digit of $2^a$ is 2 or 8. Added by 5, we got 7 or 3. There is no integer $c$ such that the last digit of $c^2$ is 7 or 3. I forgot to state that the solutions must be integer.
$endgroup$
– Agung Izzul Haq
Jan 2 at 4:32
$begingroup$
a, b, and c are integers.If $a$ is odd, then the last digit of $2^a$ is 2 or 8. Added by 5, we got 7 or 3. There is no integer $c$ such that the last digit of $c^2$ is 7 or 3. I forgot to state that the solutions must be integer.
$endgroup$
– Agung Izzul Haq
Jan 2 at 4:32
$begingroup$
Got it. My fault.
$endgroup$
– Randall
Jan 2 at 4:33
$begingroup$
Got it. My fault.
$endgroup$
– Randall
Jan 2 at 4:33
$begingroup$
@Randall: Took me a second to see it too!
$endgroup$
– JavaMan
Jan 2 at 4:33
$begingroup$
@Randall: Took me a second to see it too!
$endgroup$
– JavaMan
Jan 2 at 4:33
6
6
$begingroup$
Well, you could use Mihăilescu's theorem.
$endgroup$
– Robert Israel
Jan 2 at 4:38
$begingroup$
Well, you could use Mihăilescu's theorem.
$endgroup$
– Robert Israel
Jan 2 at 4:38
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Continue from your work, we solve $$1+2^{n+1} = 5^b$$
This special case of Catalan equation is easy: $2$ is a primitive root of $25$, hence a primitive root for any $5^b$. Therefore
$$2^{n+1} equiv -1 pmod{5^b}$$
implies $n+1$ is divisible by $varphi(5^b)/2 = 2(5^{b-1})$, so $n+1 geq 2(5^{b-1})$.
Hence $$5^b geq 1+2^{2(5^{b-1})}$$
the RHS grows far faster than LHS, the inequality only holds when $b=1$.
Therefore all solution to your original equation $2^a+5^b=c^2$ is $(a,b,c)=(2,1,3)$.
answered Jan 2 at 6:32
piscopisco
11.6k21742
$endgroup$
$begingroup$
Another approach. Thanks. But, could you explain why "$2$ is a primitive root of $25$" implies "$2$ is primitive root for any $5^b$"?
$endgroup$
– Agung Izzul Haq
Jan 2 at 9:00
$begingroup$
@AgungIzzulHaq The following result is well-known: if $b$ is a primitive root for $p^2$, then $b$ is a primitive root for any $p^n$, $ngeq 2$.
$endgroup$
– pisco
Jan 3 at 14:41
add a comment |
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$begingroup$
Continue from your work, we solve $$1+2^{n+1} = 5^b$$
This special case of Catalan equation is easy: $2$ is a primitive root of $25$, hence a primitive root for any $5^b$. Therefore
$$2^{n+1} equiv -1 pmod{5^b}$$
implies $n+1$ is divisible by $varphi(5^b)/2 = 2(5^{b-1})$, so $n+1 geq 2(5^{b-1})$.
Hence $$5^b geq 1+2^{2(5^{b-1})}$$
the RHS grows far faster than LHS, the inequality only holds when $b=1$.
Therefore all solution to your original equation $2^a+5^b=c^2$ is $(a,b,c)=(2,1,3)$.
answered Jan 2 at 6:32
piscopisco
11.6k21742
$endgroup$
$begingroup$
Another approach. Thanks. But, could you explain why "$2$ is a primitive root of $25$" implies "$2$ is primitive root for any $5^b$"?
$endgroup$
– Agung Izzul Haq
Jan 2 at 9:00
$begingroup$
@AgungIzzulHaq The following result is well-known: if $b$ is a primitive root for $p^2$, then $b$ is a primitive root for any $p^n$, $ngeq 2$.
$endgroup$
– pisco
Jan 3 at 14:41
add a comment |
$begingroup$
Continue from your work, we solve $$1+2^{n+1} = 5^b$$
This special case of Catalan equation is easy: $2$ is a primitive root of $25$, hence a primitive root for any $5^b$. Therefore
$$2^{n+1} equiv -1 pmod{5^b}$$
implies $n+1$ is divisible by $varphi(5^b)/2 = 2(5^{b-1})$, so $n+1 geq 2(5^{b-1})$.
Hence $$5^b geq 1+2^{2(5^{b-1})}$$
the RHS grows far faster than LHS, the inequality only holds when $b=1$.
Therefore all solution to your original equation $2^a+5^b=c^2$ is $(a,b,c)=(2,1,3)$.
answered Jan 2 at 6:32
piscopisco
11.6k21742
$endgroup$
$begingroup$
Another approach. Thanks. But, could you explain why "$2$ is a primitive root of $25$" implies "$2$ is primitive root for any $5^b$"?
$endgroup$
– Agung Izzul Haq
Jan 2 at 9:00
$begingroup$
@AgungIzzulHaq The following result is well-known: if $b$ is a primitive root for $p^2$, then $b$ is a primitive root for any $p^n$, $ngeq 2$.
$endgroup$
– pisco
Jan 3 at 14:41
add a comment |
$begingroup$
Continue from your work, we solve $$1+2^{n+1} = 5^b$$
This special case of Catalan equation is easy: $2$ is a primitive root of $25$, hence a primitive root for any $5^b$. Therefore
$$2^{n+1} equiv -1 pmod{5^b}$$
implies $n+1$ is divisible by $varphi(5^b)/2 = 2(5^{b-1})$, so $n+1 geq 2(5^{b-1})$.
Hence $$5^b geq 1+2^{2(5^{b-1})}$$
the RHS grows far faster than LHS, the inequality only holds when $b=1$.
Therefore all solution to your original equation $2^a+5^b=c^2$ is $(a,b,c)=(2,1,3)$.
answered Jan 2 at 6:32
piscopisco
11.6k21742
$endgroup$
Continue from your work, we solve $$1+2^{n+1} = 5^b$$
This special case of Catalan equation is easy: $2$ is a primitive root of $25$, hence a primitive root for any $5^b$. Therefore
$$2^{n+1} equiv -1 pmod{5^b}$$
implies $n+1$ is divisible by $varphi(5^b)/2 = 2(5^{b-1})$, so $n+1 geq 2(5^{b-1})$.
Hence $$5^b geq 1+2^{2(5^{b-1})}$$
the RHS grows far faster than LHS, the inequality only holds when $b=1$.
Therefore all solution to your original equation $2^a+5^b=c^2$ is $(a,b,c)=(2,1,3)$.
answered Jan 2 at 6:32
piscopisco
11.6k21742
answered Jan 2 at 6:32
piscopisco
11.6k21742
answered Jan 2 at 6:32
piscopisco
11.6k21742
answered Jan 2 at 6:32
piscopisco
11.6k21742
11.6k21742
$begingroup$
Another approach. Thanks. But, could you explain why "$2$ is a primitive root of $25$" implies "$2$ is primitive root for any $5^b$"?
$endgroup$
– Agung Izzul Haq
Jan 2 at 9:00
$begingroup$
@AgungIzzulHaq The following result is well-known: if $b$ is a primitive root for $p^2$, then $b$ is a primitive root for any $p^n$, $ngeq 2$.
$endgroup$
– pisco
Jan 3 at 14:41
add a comment |
$begingroup$
Another approach. Thanks. But, could you explain why "$2$ is a primitive root of $25$" implies "$2$ is primitive root for any $5^b$"?
$endgroup$
– Agung Izzul Haq
Jan 2 at 9:00
$begingroup$
@AgungIzzulHaq The following result is well-known: if $b$ is a primitive root for $p^2$, then $b$ is a primitive root for any $p^n$, $ngeq 2$.
$endgroup$
– pisco
Jan 3 at 14:41
$begingroup$
Another approach. Thanks. But, could you explain why "$2$ is a primitive root of $25$" implies "$2$ is primitive root for any $5^b$"?
$endgroup$
– Agung Izzul Haq
Jan 2 at 9:00
$begingroup$
Another approach. Thanks. But, could you explain why "$2$ is a primitive root of $25$" implies "$2$ is primitive root for any $5^b$"?
$endgroup$
– Agung Izzul Haq
Jan 2 at 9:00
$begingroup$
@AgungIzzulHaq The following result is well-known: if $b$ is a primitive root for $p^2$, then $b$ is a primitive root for any $p^n$, $ngeq 2$.
$endgroup$
– pisco
Jan 3 at 14:41
$begingroup$
@AgungIzzulHaq The following result is well-known: if $b$ is a primitive root for $p^2$, then $b$ is a primitive root for any $p^n$, $ngeq 2$.
$endgroup$
– pisco
Jan 3 at 14:41
add a comment |
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$begingroup$
@Randall: If $a$ is odd, then the solution fails mod $4$.
$endgroup$
– JavaMan
Jan 2 at 4:32
1
$begingroup$
a, b, and c are integers.If $a$ is odd, then the last digit of $2^a$ is 2 or 8. Added by 5, we got 7 or 3. There is no integer $c$ such that the last digit of $c^2$ is 7 or 3. I forgot to state that the solutions must be integer.
$endgroup$
– Agung Izzul Haq
Jan 2 at 4:32
$begingroup$
Got it. My fault.
$endgroup$
– Randall
Jan 2 at 4:33
$begingroup$
@Randall: Took me a second to see it too!
$endgroup$
– JavaMan
Jan 2 at 4:33
6
$begingroup$
Well, you could use Mihăilescu's theorem.
$endgroup$
– Robert Israel
Jan 2 at 4:38