Differences between numbers of the form abcdef and acdef which are perfect square.
$541456-51456=700^2$ 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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$541456-51456=700^2$ 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
add a comment |
$541456-51456=700^2$ 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
$541456-51456=700^2$ 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
asked Nov 12 '18 at 10:43
user613967
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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
answered Nov 12 '18 at 10:46
lab bhattacharjee
223k15156274
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Of course,
$$
begin{aligned}
overline{abcdef} - overline{acdef}
&=
overline{ab0000} - overline{a0000}
\
&=10000cdot(overline{ab} - overline{a})
\
&=100^2cdot(9a+b) .
end{aligned}
$$
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 Nov 12 '18 at 10:56
dan_fulea
6,2301312
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2 Answers
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2 Answers
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active
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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
answered Nov 12 '18 at 10:46
lab bhattacharjee
223k15156274
add a comment |
$$abcdef-acdef=10^4b+10^5a-(10^4a)=10^4(9a+b)$$
So, all we need is $9a+b$ to be perfect square
answered Nov 12 '18 at 10:46
lab bhattacharjee
223k15156274
add a comment |
$$abcdef-acdef=10^4b+10^5a-(10^4a)=10^4(9a+b)$$
So, all we need is $9a+b$ to be perfect square
answered Nov 12 '18 at 10:46
lab bhattacharjee
223k15156274
$$abcdef-acdef=10^4b+10^5a-(10^4a)=10^4(9a+b)$$
So, all we need is $9a+b$ to be perfect square
answered Nov 12 '18 at 10:46
lab bhattacharjee
223k15156274
answered Nov 12 '18 at 10:46
lab bhattacharjee
223k15156274
answered Nov 12 '18 at 10:46
lab bhattacharjee
223k15156274
answered Nov 12 '18 at 10:46
lab bhattacharjee
223k15156274
223k15156274
add a comment |
add a comment |
Of course,
$$
begin{aligned}
overline{abcdef} - overline{acdef}
&=
overline{ab0000} - overline{a0000}
\
&=10000cdot(overline{ab} - overline{a})
\
&=100^2cdot(9a+b) .
end{aligned}
$$
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 Nov 12 '18 at 10:56
dan_fulea
6,2301312
add a comment |
Of course,
$$
begin{aligned}
overline{abcdef} - overline{acdef}
&=
overline{ab0000} - overline{a0000}
\
&=10000cdot(overline{ab} - overline{a})
\
&=100^2cdot(9a+b) .
end{aligned}
$$
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 Nov 12 '18 at 10:56
dan_fulea
6,2301312
add a comment |
Of course,
$$
begin{aligned}
overline{abcdef} - overline{acdef}
&=
overline{ab0000} - overline{a0000}
\
&=10000cdot(overline{ab} - overline{a})
\
&=100^2cdot(9a+b) .
end{aligned}
$$
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 Nov 12 '18 at 10:56
dan_fulea
6,2301312
Of course,
$$
begin{aligned}
overline{abcdef} - overline{acdef}
&=
overline{ab0000} - overline{a0000}
\
&=10000cdot(overline{ab} - overline{a})
\
&=100^2cdot(9a+b) .
end{aligned}
$$
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 Nov 12 '18 at 10:56
dan_fulea
6,2301312
answered Nov 12 '18 at 10:56
dan_fulea
6,2301312
answered Nov 12 '18 at 10:56
dan_fulea
6,2301312
answered Nov 12 '18 at 10:56
dan_fulea
6,2301312
6,2301312
add a comment |
add a comment |
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