Prove that there are no natural numbers $x$ whose digits are $0$ or $2$, such that $x$ is a perfect square.
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Prove that there are no natural numbers $x$ whose digits are $0$ or $2$, such that $x$ is a perfect square. I need some help here. I thought starting with $x = n * 10^k$ where $10^k$ represents the number of the zeros at the end of the number and $n$ is the group of digits which end with $2$ could help, but it didn't. Can anyone help?
linear-algebra number-theory elementary-number-theory
asked Jan 18 at 1:11
Wolf M.Wolf M.
1097
$endgroup$
add a comment |
$begingroup$
Prove that there are no natural numbers $x$ whose digits are $0$ or $2$, such that $x$ is a perfect square. I need some help here. I thought starting with $x = n * 10^k$ where $10^k$ represents the number of the zeros at the end of the number and $n$ is the group of digits which end with $2$ could help, but it didn't. Can anyone help?
linear-algebra number-theory elementary-number-theory
asked Jan 18 at 1:11
Wolf M.Wolf M.
1097
$endgroup$
1
$begingroup$
It might be more fruitful to decompose $y = n cdot 10^k$ such that $10$ does not divide $n$, where $x = y^2$.
$endgroup$
– Daniel Schepler
Jan 18 at 1:29
7
$begingroup$
The last non-zero digit of a perfect square is always 1, 4, 5, 6, or 9.
$endgroup$
– TonyK
Jan 18 at 1:31
add a comment |
$begingroup$
Prove that there are no natural numbers $x$ whose digits are $0$ or $2$, such that $x$ is a perfect square. I need some help here. I thought starting with $x = n * 10^k$ where $10^k$ represents the number of the zeros at the end of the number and $n$ is the group of digits which end with $2$ could help, but it didn't. Can anyone help?
linear-algebra number-theory elementary-number-theory
asked Jan 18 at 1:11
Wolf M.Wolf M.
1097
$endgroup$
Prove that there are no natural numbers $x$ whose digits are $0$ or $2$, such that $x$ is a perfect square. I need some help here. I thought starting with $x = n * 10^k$ where $10^k$ represents the number of the zeros at the end of the number and $n$ is the group of digits which end with $2$ could help, but it didn't. Can anyone help?
linear-algebra number-theory elementary-number-theory
linear-algebra number-theory elementary-number-theory
asked Jan 18 at 1:11
Wolf M.Wolf M.
1097
asked Jan 18 at 1:11
Wolf M.Wolf M.
1097
asked Jan 18 at 1:11
Wolf M.Wolf M.
1097
asked Jan 18 at 1:11
Wolf M.Wolf M.
1097
asked Jan 18 at 1:11
Wolf M.Wolf M.
1097
1097
1
$begingroup$
It might be more fruitful to decompose $y = n cdot 10^k$ such that $10$ does not divide $n$, where $x = y^2$.
$endgroup$
– Daniel Schepler
Jan 18 at 1:29
7
$begingroup$
The last non-zero digit of a perfect square is always 1, 4, 5, 6, or 9.
$endgroup$
– TonyK
Jan 18 at 1:31
add a comment |
1
$begingroup$
It might be more fruitful to decompose $y = n cdot 10^k$ such that $10$ does not divide $n$, where $x = y^2$.
$endgroup$
– Daniel Schepler
Jan 18 at 1:29
7
$begingroup$
The last non-zero digit of a perfect square is always 1, 4, 5, 6, or 9.
$endgroup$
– TonyK
Jan 18 at 1:31
1
1
$begingroup$
It might be more fruitful to decompose $y = n cdot 10^k$ such that $10$ does not divide $n$, where $x = y^2$.
$endgroup$
– Daniel Schepler
Jan 18 at 1:29
$begingroup$
It might be more fruitful to decompose $y = n cdot 10^k$ such that $10$ does not divide $n$, where $x = y^2$.
$endgroup$
– Daniel Schepler
Jan 18 at 1:29
7
7
$begingroup$
The last non-zero digit of a perfect square is always 1, 4, 5, 6, or 9.
$endgroup$
– TonyK
Jan 18 at 1:31
$begingroup$
The last non-zero digit of a perfect square is always 1, 4, 5, 6, or 9.
$endgroup$
– TonyK
Jan 18 at 1:31
add a comment |
1 Answer
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Dividing by an appropriate power of $100$ we can ensure that the final two digits are not both $0$. But a simple search (or congruence argument) shows that none of $2,20,22$ are squares $pmod {100}$.
answered Jan 18 at 1:15
lulululu
43.7k25081
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add a comment |
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$begingroup$
Dividing by an appropriate power of $100$ we can ensure that the final two digits are not both $0$. But a simple search (or congruence argument) shows that none of $2,20,22$ are squares $pmod {100}$.
answered Jan 18 at 1:15
lulululu
43.7k25081
$endgroup$
add a comment |
$begingroup$
Dividing by an appropriate power of $100$ we can ensure that the final two digits are not both $0$. But a simple search (or congruence argument) shows that none of $2,20,22$ are squares $pmod {100}$.
answered Jan 18 at 1:15
lulululu
43.7k25081
$endgroup$
add a comment |
$begingroup$
Dividing by an appropriate power of $100$ we can ensure that the final two digits are not both $0$. But a simple search (or congruence argument) shows that none of $2,20,22$ are squares $pmod {100}$.
answered Jan 18 at 1:15
lulululu
43.7k25081
$endgroup$
Dividing by an appropriate power of $100$ we can ensure that the final two digits are not both $0$. But a simple search (or congruence argument) shows that none of $2,20,22$ are squares $pmod {100}$.
answered Jan 18 at 1:15
lulululu
43.7k25081
edited Jan 18 at 2:27
answered Jan 18 at 1:15
lulululu
43.7k25081
answered Jan 18 at 1:15
lulululu
43.7k25081
answered Jan 18 at 1:15
lulululu
43.7k25081
43.7k25081
add a comment |
add a comment |
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$begingroup$
It might be more fruitful to decompose $y = n cdot 10^k$ such that $10$ does not divide $n$, where $x = y^2$.
$endgroup$
– Daniel Schepler
Jan 18 at 1:29
7
$begingroup$
The last non-zero digit of a perfect square is always 1, 4, 5, 6, or 9.
$endgroup$
– TonyK
Jan 18 at 1:31