Differences between numbers of the form abcdef and acdef which are perfect square.
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$541456-51456=700^2$ is the example I found. Are there other examples of numbers of the form abcdef such that abcdef-acdef is a perfect square?
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$541456-51456=700^2$ is the example I found. Are there other examples of numbers of the form abcdef such that abcdef-acdef is a perfect square?
number-theory
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up vote
2
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up vote
2
down vote
favorite
$541456-51456=700^2$ is the example I found. Are there other examples of numbers of the form abcdef such that abcdef-acdef is a perfect square?
number-theory
New contributor
$541456-51456=700^2$ is the example I found. Are there other examples of numbers of the form abcdef such that abcdef-acdef is a perfect square?
number-theory
number-theory
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New contributor
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asked 14 hours ago
homunculus
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393
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2 Answers
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4
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$$abcdef-acdef=10^4b+10^5a-(10^4a)=10^4(9a+b)$$
So, all we need is $9a+b$ to be perfect square
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1
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Of course,
$$
beginaligned
overlineabcdef - overlineacdef
&=
overlineab0000 - overlinea0000
\
&=10000cdot(overlineab - overlinea)
\
&=100^2cdot(9a+b) .
endaligned
$$
So we can choose $c,d,e,f$ arbitrarily, and for $a,b$ there are the following chances:
for a in [1..9]:
for b in [0..9]:
ab_a = 9*a + b
if ab_a.is_square():
print ( "a=%s b=%s %s-%s = %s^2"
% (a, b, 10*a+b, a, sqrt(ab_a)) )
And the results are:
a=1 b=0 10-1 = 3^2
a=1 b=7 17-1 = 4^2
a=2 b=7 27-2 = 5^2
a=3 b=9 39-3 = 6^2
a=4 b=0 40-4 = 6^2
a=5 b=4 54-5 = 7^2
a=7 b=1 71-7 = 8^2
a=8 b=9 89-8 = 9^2
a=9 b=0 90-9 = 9^2
Here i was using sage.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
$$abcdef-acdef=10^4b+10^5a-(10^4a)=10^4(9a+b)$$
So, all we need is $9a+b$ to be perfect square
add a comment |
up vote
4
down vote
$$abcdef-acdef=10^4b+10^5a-(10^4a)=10^4(9a+b)$$
So, all we need is $9a+b$ to be perfect square
add a comment |
up vote
4
down vote
up vote
4
down vote
$$abcdef-acdef=10^4b+10^5a-(10^4a)=10^4(9a+b)$$
So, all we need is $9a+b$ to be perfect square
$$abcdef-acdef=10^4b+10^5a-(10^4a)=10^4(9a+b)$$
So, all we need is $9a+b$ to be perfect square
answered 14 hours ago
lab bhattacharjee
219k14153268
219k14153268
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up vote
1
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Of course,
$$
beginaligned
overlineabcdef - overlineacdef
&=
overlineab0000 - overlinea0000
\
&=10000cdot(overlineab - overlinea)
\
&=100^2cdot(9a+b) .
endaligned
$$
So we can choose $c,d,e,f$ arbitrarily, and for $a,b$ there are the following chances:
for a in [1..9]:
for b in [0..9]:
ab_a = 9*a + b
if ab_a.is_square():
print ( "a=%s b=%s %s-%s = %s^2"
% (a, b, 10*a+b, a, sqrt(ab_a)) )
And the results are:
a=1 b=0 10-1 = 3^2
a=1 b=7 17-1 = 4^2
a=2 b=7 27-2 = 5^2
a=3 b=9 39-3 = 6^2
a=4 b=0 40-4 = 6^2
a=5 b=4 54-5 = 7^2
a=7 b=1 71-7 = 8^2
a=8 b=9 89-8 = 9^2
a=9 b=0 90-9 = 9^2
Here i was using sage.
add a comment |
up vote
1
down vote
Of course,
$$
beginaligned
overlineabcdef - overlineacdef
&=
overlineab0000 - overlinea0000
\
&=10000cdot(overlineab - overlinea)
\
&=100^2cdot(9a+b) .
endaligned
$$
So we can choose $c,d,e,f$ arbitrarily, and for $a,b$ there are the following chances:
for a in [1..9]:
for b in [0..9]:
ab_a = 9*a + b
if ab_a.is_square():
print ( "a=%s b=%s %s-%s = %s^2"
% (a, b, 10*a+b, a, sqrt(ab_a)) )
And the results are:
a=1 b=0 10-1 = 3^2
a=1 b=7 17-1 = 4^2
a=2 b=7 27-2 = 5^2
a=3 b=9 39-3 = 6^2
a=4 b=0 40-4 = 6^2
a=5 b=4 54-5 = 7^2
a=7 b=1 71-7 = 8^2
a=8 b=9 89-8 = 9^2
a=9 b=0 90-9 = 9^2
Here i was using sage.
add a comment |
up vote
1
down vote
up vote
1
down vote
Of course,
$$
beginaligned
overlineabcdef - overlineacdef
&=
overlineab0000 - overlinea0000
\
&=10000cdot(overlineab - overlinea)
\
&=100^2cdot(9a+b) .
endaligned
$$
So we can choose $c,d,e,f$ arbitrarily, and for $a,b$ there are the following chances:
for a in [1..9]:
for b in [0..9]:
ab_a = 9*a + b
if ab_a.is_square():
print ( "a=%s b=%s %s-%s = %s^2"
% (a, b, 10*a+b, a, sqrt(ab_a)) )
And the results are:
a=1 b=0 10-1 = 3^2
a=1 b=7 17-1 = 4^2
a=2 b=7 27-2 = 5^2
a=3 b=9 39-3 = 6^2
a=4 b=0 40-4 = 6^2
a=5 b=4 54-5 = 7^2
a=7 b=1 71-7 = 8^2
a=8 b=9 89-8 = 9^2
a=9 b=0 90-9 = 9^2
Here i was using sage.
Of course,
$$
beginaligned
overlineabcdef - overlineacdef
&=
overlineab0000 - overlinea0000
\
&=10000cdot(overlineab - overlinea)
\
&=100^2cdot(9a+b) .
endaligned
$$
So we can choose $c,d,e,f$ arbitrarily, and for $a,b$ there are the following chances:
for a in [1..9]:
for b in [0..9]:
ab_a = 9*a + b
if ab_a.is_square():
print ( "a=%s b=%s %s-%s = %s^2"
% (a, b, 10*a+b, a, sqrt(ab_a)) )
And the results are:
a=1 b=0 10-1 = 3^2
a=1 b=7 17-1 = 4^2
a=2 b=7 27-2 = 5^2
a=3 b=9 39-3 = 6^2
a=4 b=0 40-4 = 6^2
a=5 b=4 54-5 = 7^2
a=7 b=1 71-7 = 8^2
a=8 b=9 89-8 = 9^2
a=9 b=0 90-9 = 9^2
Here i was using sage.
answered 14 hours ago
dan_fulea
5,7251312
5,7251312
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