Find all positive integers $n$ such that $12n-119$ and $75n-539$ are both perfect squares. [closed] The 2019 Stack Overflow Developer Survey Results Are InFinding pairs of integers such that $x^2+3y$ and $y^2+3x$ are both perfect squaresFinding all integers such that $a^2+4b^2 , 4a^2+b^2$ are both perfect squaresFind integers $a$ and $b$ such that $a^5b+3$ and $ab^5+3$ are both perfect cubes of integers?Prove that both $4m^2+17n^2$ and $4n^2+17m^2$ cannot be perfect squares for positive integers $m$ and $n$.There are two integers whose sum and difference are perfect squaresHow many positive integers $n$ are there such that $2n$ and $3n$ both perfect squares?Show that if $x,y,z$ are positive integers, then $(xy + 1)(yz + 1)(zx + 1)$ is a perfect square iff $xy +1, yz +1, zx+1$ are all perfect squares.If $ab+1$ is perfect square there is a $k$ such that $ak+1$ and $bk+1$ are perfect squaresBoth $m^2 + n^2 $ and $m^2-n^2$ are not perfect squares.Suppose a, b are integers such that both 2a+3b and 3a-2b are the squares of positive integers. What is the smallest possible values of these squares?
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Find all positive integers $n$ such that $12n-119$ and $75n-539$ are both perfect squares. [closed]
The 2019 Stack Overflow Developer Survey Results Are InFinding pairs of integers such that $x^2+3y$ and $y^2+3x$ are both perfect squaresFinding all integers such that $a^2+4b^2 , 4a^2+b^2$ are both perfect squaresFind integers $a$ and $b$ such that $a^5b+3$ and $ab^5+3$ are both perfect cubes of integers?Prove that both $4m^2+17n^2$ and $4n^2+17m^2$ cannot be perfect squares for positive integers $m$ and $n$.There are two integers whose sum and difference are perfect squaresHow many positive integers $n$ are there such that $2n$ and $3n$ both perfect squares?Show that if $x,y,z$ are positive integers, then $(xy + 1)(yz + 1)(zx + 1)$ is a perfect square iff $xy +1, yz +1, zx+1$ are all perfect squares.If $ab+1$ is perfect square there is a $k$ such that $ak+1$ and $bk+1$ are perfect squaresBoth $m^2 + n^2 $ and $m^2-n^2$ are not perfect squares.Suppose a, b are integers such that both 2a+3b and 3a-2b are the squares of positive integers. What is the smallest possible values of these squares?
$begingroup$
Find all positive integers $n$ such that $12n-119$ and $75n-539$ are both perfect squares.
elementary-number-theory
edited Mar 30 at 11:04
Mars Plastic
1,455122
asked Mar 30 at 9:57
Anson ChanAnson Chan
162
$endgroup$
closed as off-topic by Sil, GNUSupporter 8964民主女神 地下教會, Javi, mrtaurho, John Omielan Mar 30 at 16:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Sil, GNUSupporter 8964民主女神 地下教會, Javi, mrtaurho, John Omielan
add a comment |
$begingroup$
Find all positive integers $n$ such that $12n-119$ and $75n-539$ are both perfect squares.
elementary-number-theory
edited Mar 30 at 11:04
Mars Plastic
1,455122
asked Mar 30 at 9:57
Anson ChanAnson Chan
162
$endgroup$
closed as off-topic by Sil, GNUSupporter 8964民主女神 地下教會, Javi, mrtaurho, John Omielan Mar 30 at 16:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Sil, GNUSupporter 8964民主女神 地下教會, Javi, mrtaurho, John Omielan
add a comment |
$begingroup$
Find all positive integers $n$ such that $12n-119$ and $75n-539$ are both perfect squares.
elementary-number-theory
edited Mar 30 at 11:04
Mars Plastic
1,455122
asked Mar 30 at 9:57
Anson ChanAnson Chan
162
$endgroup$
Find all positive integers $n$ such that $12n-119$ and $75n-539$ are both perfect squares.
elementary-number-theory
elementary-number-theory
edited Mar 30 at 11:04
Mars Plastic
1,455122
asked Mar 30 at 9:57
Anson ChanAnson Chan
162
edited Mar 30 at 11:04
Mars Plastic
1,455122
asked Mar 30 at 9:57
Anson ChanAnson Chan
162
edited Mar 30 at 11:04
Mars Plastic
1,455122
edited Mar 30 at 11:04
Mars Plastic
1,455122
edited Mar 30 at 11:04
Mars Plastic
1,455122
1,455122
asked Mar 30 at 9:57
Anson ChanAnson Chan
162
asked Mar 30 at 9:57
Anson ChanAnson Chan
162
asked Mar 30 at 9:57
Anson ChanAnson Chan
162
162
closed as off-topic by Sil, GNUSupporter 8964民主女神 地下教會, Javi, mrtaurho, John Omielan Mar 30 at 16:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Sil, GNUSupporter 8964民主女神 地下教會, Javi, mrtaurho, John Omielan
closed as off-topic by Sil, GNUSupporter 8964民主女神 地下教會, Javi, mrtaurho, John Omielan Mar 30 at 16:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Sil, GNUSupporter 8964民主女神 地下教會, Javi, mrtaurho, John Omielan
add a comment |
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1 Answer
1
active
oldest
votes
$begingroup$
Let
$12n-119=a^2...(1)$
$75n-539=b^2...(2)$where $a,b$ are positive integers.
By $25cdot (1)-4cdot(2)$ the $n$ cancells out
$(5a+2b)(5a-2b)= -819$
$(5a+2b)(2b-5a)= 819$
As both algebraic terms are integers, they are both pair of factors of $819$
$implies 5a+2b=1 2b-5a=819$
$5a+2b=3 2b-5a=273$
(etc.)
$5a+2b=819 2b-5a=1$
By finding integer solutions of these simultaneous equations, we have:
$a=5,b=19$
$a=11,b=31$
$a=27,b=69$
But the last pair of solution won't give n as an integer.
$implies n=12, 20$
edited Mar 30 at 16:34
Dr. Mathva
3,493630
answered Mar 30 at 10:44
user659210user659210
192
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let
$12n-119=a^2...(1)$
$75n-539=b^2...(2)$where $a,b$ are positive integers.
By $25cdot (1)-4cdot(2)$ the $n$ cancells out
$(5a+2b)(5a-2b)= -819$
$(5a+2b)(2b-5a)= 819$
As both algebraic terms are integers, they are both pair of factors of $819$
$implies 5a+2b=1 2b-5a=819$
$5a+2b=3 2b-5a=273$
(etc.)
$5a+2b=819 2b-5a=1$
By finding integer solutions of these simultaneous equations, we have:
$a=5,b=19$
$a=11,b=31$
$a=27,b=69$
But the last pair of solution won't give n as an integer.
$implies n=12, 20$
edited Mar 30 at 16:34
Dr. Mathva
3,493630
answered Mar 30 at 10:44
user659210user659210
192
$endgroup$
add a comment |
$begingroup$
Let
$12n-119=a^2...(1)$
$75n-539=b^2...(2)$where $a,b$ are positive integers.
By $25cdot (1)-4cdot(2)$ the $n$ cancells out
$(5a+2b)(5a-2b)= -819$
$(5a+2b)(2b-5a)= 819$
As both algebraic terms are integers, they are both pair of factors of $819$
$implies 5a+2b=1 2b-5a=819$
$5a+2b=3 2b-5a=273$
(etc.)
$5a+2b=819 2b-5a=1$
By finding integer solutions of these simultaneous equations, we have:
$a=5,b=19$
$a=11,b=31$
$a=27,b=69$
But the last pair of solution won't give n as an integer.
$implies n=12, 20$
edited Mar 30 at 16:34
Dr. Mathva
3,493630
answered Mar 30 at 10:44
user659210user659210
192
$endgroup$
add a comment |
$begingroup$
Let
$12n-119=a^2...(1)$
$75n-539=b^2...(2)$where $a,b$ are positive integers.
By $25cdot (1)-4cdot(2)$ the $n$ cancells out
$(5a+2b)(5a-2b)= -819$
$(5a+2b)(2b-5a)= 819$
As both algebraic terms are integers, they are both pair of factors of $819$
$implies 5a+2b=1 2b-5a=819$
$5a+2b=3 2b-5a=273$
(etc.)
$5a+2b=819 2b-5a=1$
By finding integer solutions of these simultaneous equations, we have:
$a=5,b=19$
$a=11,b=31$
$a=27,b=69$
But the last pair of solution won't give n as an integer.
$implies n=12, 20$
edited Mar 30 at 16:34
Dr. Mathva
3,493630
answered Mar 30 at 10:44
user659210user659210
192
$endgroup$
Let
$12n-119=a^2...(1)$
$75n-539=b^2...(2)$where $a,b$ are positive integers.
By $25cdot (1)-4cdot(2)$ the $n$ cancells out
$(5a+2b)(5a-2b)= -819$
$(5a+2b)(2b-5a)= 819$
As both algebraic terms are integers, they are both pair of factors of $819$
$implies 5a+2b=1 2b-5a=819$
$5a+2b=3 2b-5a=273$
(etc.)
$5a+2b=819 2b-5a=1$
By finding integer solutions of these simultaneous equations, we have:
$a=5,b=19$
$a=11,b=31$
$a=27,b=69$
But the last pair of solution won't give n as an integer.
$implies n=12, 20$
edited Mar 30 at 16:34
Dr. Mathva
3,493630
answered Mar 30 at 10:44
user659210user659210
192
edited Mar 30 at 16:34
Dr. Mathva
3,493630
edited Mar 30 at 16:34
Dr. Mathva
3,493630
edited Mar 30 at 16:34
Dr. Mathva
3,493630
3,493630
answered Mar 30 at 10:44
user659210user659210
192
answered Mar 30 at 10:44
user659210user659210
192
answered Mar 30 at 10:44
user659210user659210
192
192
add a comment |
add a comment |
