Why does my book give a few extra solutions?
$begingroup$
I have the following question from INMO 2004, Find all pairs of integers (a,b) satisfying the equation, where $p$ is prime
$$a^2 + 3ab + 2p(a+b) + p^2 = 0$$
for which I proceed this way:
$$(a+b+p)^2 = b^2 - ab$$
which means $b^2 - ab$ is a perfect square. Let $gcd(a,b) = t$ where $a = st$ and $b = rt$ where $gcd(r,s) = 1$. which means $t^2r(r-s)$ is a perfect square however $gcd(r,r-s) = 1$ which means both $r$ and $r-s$ are perfect squares. Let $r = alpha^2$and $s=alpha^2 - beta^2$ which reduces the above equation to $$(a+b+p)^2 = alpha^2beta^2t^2$$
which implies either $$a + b + p = alphabeta t$$or
$$a+b+p = -alphabeta t$$
again substituting a and b as above in terms of $alpha,beta,t$ I get $$p = t(beta - alpha)(2alpha + beta)$$ in first case and $$p = t(alpha + beta)(-2alpha + beta)$$ in the second case. Since $p$ is a prime, we have two terms of the above expression to be one. Upon solving I can get $3$ solutions, two of which are obtained specifically only when $p$ is congruent to $1$ (mod $3)$. Similarly, upon solving the second case above, I again end up with same set of solutions for $(a,b)$ again specifically for $p$ congruent to $1$ (mod $3$).
However my book says solutions exist even for $p$ congruent to $-1$ mod $3$. However They aren't coming up in my solution, why is it so? Have I missed any case?
number-theory proof-verification contest-math
edited Jan 2 at 8:57
Sauhard Sharma
953318
asked Jan 2 at 6:53
saisanjeevsaisanjeev
1,071312
$endgroup$
add a comment |
$begingroup$
I have the following question from INMO 2004, Find all pairs of integers (a,b) satisfying the equation, where $p$ is prime
$$a^2 + 3ab + 2p(a+b) + p^2 = 0$$
for which I proceed this way:
$$(a+b+p)^2 = b^2 - ab$$
which means $b^2 - ab$ is a perfect square. Let $gcd(a,b) = t$ where $a = st$ and $b = rt$ where $gcd(r,s) = 1$. which means $t^2r(r-s)$ is a perfect square however $gcd(r,r-s) = 1$ which means both $r$ and $r-s$ are perfect squares. Let $r = alpha^2$and $s=alpha^2 - beta^2$ which reduces the above equation to $$(a+b+p)^2 = alpha^2beta^2t^2$$
which implies either $$a + b + p = alphabeta t$$or
$$a+b+p = -alphabeta t$$
again substituting a and b as above in terms of $alpha,beta,t$ I get $$p = t(beta - alpha)(2alpha + beta)$$ in first case and $$p = t(alpha + beta)(-2alpha + beta)$$ in the second case. Since $p$ is a prime, we have two terms of the above expression to be one. Upon solving I can get $3$ solutions, two of which are obtained specifically only when $p$ is congruent to $1$ (mod $3)$. Similarly, upon solving the second case above, I again end up with same set of solutions for $(a,b)$ again specifically for $p$ congruent to $1$ (mod $3$).
However my book says solutions exist even for $p$ congruent to $-1$ mod $3$. However They aren't coming up in my solution, why is it so? Have I missed any case?
number-theory proof-verification contest-math
edited Jan 2 at 8:57
Sauhard Sharma
953318
asked Jan 2 at 6:53
saisanjeevsaisanjeev
1,071312
$endgroup$
3
$begingroup$
There is a possibility that either $a$ or $b$ are $0$. For example, $a=-p$ and $b=0$ is a solution. So definitely there are solutions for $p equiv pm 1 pmod{3}$. When you say that two terms should be $1$ because $p$ is a prime. There there is one more possibility that they are both $-1$.
$endgroup$
– Anurag A
Jan 2 at 7:15
$begingroup$
oh yeah that's perfect, solves my problem completely. Thanks a ton
$endgroup$
– saisanjeev
Jan 2 at 8:05
add a comment |
$begingroup$
I have the following question from INMO 2004, Find all pairs of integers (a,b) satisfying the equation, where $p$ is prime
$$a^2 + 3ab + 2p(a+b) + p^2 = 0$$
for which I proceed this way:
$$(a+b+p)^2 = b^2 - ab$$
which means $b^2 - ab$ is a perfect square. Let $gcd(a,b) = t$ where $a = st$ and $b = rt$ where $gcd(r,s) = 1$. which means $t^2r(r-s)$ is a perfect square however $gcd(r,r-s) = 1$ which means both $r$ and $r-s$ are perfect squares. Let $r = alpha^2$and $s=alpha^2 - beta^2$ which reduces the above equation to $$(a+b+p)^2 = alpha^2beta^2t^2$$
which implies either $$a + b + p = alphabeta t$$or
$$a+b+p = -alphabeta t$$
again substituting a and b as above in terms of $alpha,beta,t$ I get $$p = t(beta - alpha)(2alpha + beta)$$ in first case and $$p = t(alpha + beta)(-2alpha + beta)$$ in the second case. Since $p$ is a prime, we have two terms of the above expression to be one. Upon solving I can get $3$ solutions, two of which are obtained specifically only when $p$ is congruent to $1$ (mod $3)$. Similarly, upon solving the second case above, I again end up with same set of solutions for $(a,b)$ again specifically for $p$ congruent to $1$ (mod $3$).
However my book says solutions exist even for $p$ congruent to $-1$ mod $3$. However They aren't coming up in my solution, why is it so? Have I missed any case?
number-theory proof-verification contest-math
edited Jan 2 at 8:57
Sauhard Sharma
953318
asked Jan 2 at 6:53
saisanjeevsaisanjeev
1,071312
$endgroup$
I have the following question from INMO 2004, Find all pairs of integers (a,b) satisfying the equation, where $p$ is prime
$$a^2 + 3ab + 2p(a+b) + p^2 = 0$$
for which I proceed this way:
$$(a+b+p)^2 = b^2 - ab$$
which means $b^2 - ab$ is a perfect square. Let $gcd(a,b) = t$ where $a = st$ and $b = rt$ where $gcd(r,s) = 1$. which means $t^2r(r-s)$ is a perfect square however $gcd(r,r-s) = 1$ which means both $r$ and $r-s$ are perfect squares. Let $r = alpha^2$and $s=alpha^2 - beta^2$ which reduces the above equation to $$(a+b+p)^2 = alpha^2beta^2t^2$$
which implies either $$a + b + p = alphabeta t$$or
$$a+b+p = -alphabeta t$$
again substituting a and b as above in terms of $alpha,beta,t$ I get $$p = t(beta - alpha)(2alpha + beta)$$ in first case and $$p = t(alpha + beta)(-2alpha + beta)$$ in the second case. Since $p$ is a prime, we have two terms of the above expression to be one. Upon solving I can get $3$ solutions, two of which are obtained specifically only when $p$ is congruent to $1$ (mod $3)$. Similarly, upon solving the second case above, I again end up with same set of solutions for $(a,b)$ again specifically for $p$ congruent to $1$ (mod $3$).
However my book says solutions exist even for $p$ congruent to $-1$ mod $3$. However They aren't coming up in my solution, why is it so? Have I missed any case?
number-theory proof-verification contest-math
number-theory proof-verification contest-math
edited Jan 2 at 8:57
Sauhard Sharma
953318
asked Jan 2 at 6:53
saisanjeevsaisanjeev
1,071312
edited Jan 2 at 8:57
Sauhard Sharma
953318
asked Jan 2 at 6:53
saisanjeevsaisanjeev
1,071312
edited Jan 2 at 8:57
Sauhard Sharma
953318
edited Jan 2 at 8:57
Sauhard Sharma
953318
edited Jan 2 at 8:57
Sauhard Sharma
953318
953318
asked Jan 2 at 6:53
saisanjeevsaisanjeev
1,071312
asked Jan 2 at 6:53
saisanjeevsaisanjeev
1,071312
asked Jan 2 at 6:53
saisanjeevsaisanjeev
1,071312
1,071312
3
$begingroup$
There is a possibility that either $a$ or $b$ are $0$. For example, $a=-p$ and $b=0$ is a solution. So definitely there are solutions for $p equiv pm 1 pmod{3}$. When you say that two terms should be $1$ because $p$ is a prime. There there is one more possibility that they are both $-1$.
$endgroup$
– Anurag A
Jan 2 at 7:15
$begingroup$
oh yeah that's perfect, solves my problem completely. Thanks a ton
$endgroup$
– saisanjeev
Jan 2 at 8:05
add a comment |
3
$begingroup$
There is a possibility that either $a$ or $b$ are $0$. For example, $a=-p$ and $b=0$ is a solution. So definitely there are solutions for $p equiv pm 1 pmod{3}$. When you say that two terms should be $1$ because $p$ is a prime. There there is one more possibility that they are both $-1$.
$endgroup$
– Anurag A
Jan 2 at 7:15
$begingroup$
oh yeah that's perfect, solves my problem completely. Thanks a ton
$endgroup$
– saisanjeev
Jan 2 at 8:05
3
3
$begingroup$
There is a possibility that either $a$ or $b$ are $0$. For example, $a=-p$ and $b=0$ is a solution. So definitely there are solutions for $p equiv pm 1 pmod{3}$. When you say that two terms should be $1$ because $p$ is a prime. There there is one more possibility that they are both $-1$.
$endgroup$
– Anurag A
Jan 2 at 7:15
$begingroup$
There is a possibility that either $a$ or $b$ are $0$. For example, $a=-p$ and $b=0$ is a solution. So definitely there are solutions for $p equiv pm 1 pmod{3}$. When you say that two terms should be $1$ because $p$ is a prime. There there is one more possibility that they are both $-1$.
$endgroup$
– Anurag A
Jan 2 at 7:15
$begingroup$
oh yeah that's perfect, solves my problem completely. Thanks a ton
$endgroup$
– saisanjeev
Jan 2 at 8:05
$begingroup$
oh yeah that's perfect, solves my problem completely. Thanks a ton
$endgroup$
– saisanjeev
Jan 2 at 8:05
add a comment |
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3
$begingroup$
There is a possibility that either $a$ or $b$ are $0$. For example, $a=-p$ and $b=0$ is a solution. So definitely there are solutions for $p equiv pm 1 pmod{3}$. When you say that two terms should be $1$ because $p$ is a prime. There there is one more possibility that they are both $-1$.
$endgroup$
– Anurag A
Jan 2 at 7:15
$begingroup$
oh yeah that's perfect, solves my problem completely. Thanks a ton
$endgroup$
– saisanjeev
Jan 2 at 8:05