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Show that for infinitely many pairs of $(a,b)$ for $x^2+ax+b=0$ and $x^2+2ax+b=0$ have integer roots.

Proof help involving perfect squareslinear combinations of Chebyshev polynomials of first and second kindProve that there exists infinitely many pairs of relatively prime integers $(a,b)$.For which values of $n$ the sum $sum_k=1^n k^2$ is a perfect square?Show that there are infinitely many pairs (a,b) such that both $x^2+ax+b$ and $x^2+2ax+b$ have integer rootsThe number of distinct least prime factors in a sequence of consecutive integersInequalities for $operatornamerad(ntextth pentagonal number)$ and the diagonal of the regular pentagon that represents this polygonal numberWhich natural numbers are the sum of three positive perfect squares?For what $n$ does there exist $a,b > 0$ such that there are exactly $n$ primes $p$ such that $ap + b$ is a perfect square?Two quadratic functions with simillar coefficients and a square of their 'b' coefficient

0

1

$begingroup$

Show that there are infinitely many pairs $(a,b)$ of relatively prime integers (not necessarily positive) such that both quadratic equations $x^2+ax+b=0$ and $x^2+2ax+b=0$ have integer roots.

I noticed the obvious relation that $(a^2-4b)$ and $(4a^2-4b)$ both are perfect squares. Can we do anything from this? Is there any elegant and GOOD solution for this question? I failed to do anything but spending three hours.

Please help me thank you

elementary-number-theory polynomials

share|cite|improve this question

edited Mar 21 at 5:14

Sufaid Saleel

asked Mar 21 at 5:09

Sufaid SaleelSufaid Saleel

1,756829

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Technically, the expressions you have need to be four times some perfect square, some integers $m,n$ such that $(a^2 - 4b) = 4m^2, (4a^2 - 4b) = 4n^2$. That way when you take the square root, you have $2n,2m$. The quadratic formula requires you to divide by two and you want your roots to be integers. (Also the latter expression, $(4a^2 - 4b)$, can have its four factored out, so you just really need $(a^2 - b) = n^2$ for that as a result. ... Not that this is a ton of help but it's worth noting nonetheless
  $endgroup$
  – Eevee Trainer
  Mar 21 at 5:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @EeveeTrainer The division by $2$ isn't a problem - because we also have that linear term $-a$ before we divide by $2$. If $sqrta^2-4b$ is an odd integer, $-a+sqrta^2-4b$ and $-a-sqrta^2-4b$ are both even.
  $endgroup$
  – jmerry
  Mar 21 at 5:31
  

  
  
  
  

add a comment | 

0

1

$begingroup$

Show that there are infinitely many pairs $(a,b)$ of relatively prime integers (not necessarily positive) such that both quadratic equations $x^2+ax+b=0$ and $x^2+2ax+b=0$ have integer roots.

I noticed the obvious relation that $(a^2-4b)$ and $(4a^2-4b)$ both are perfect squares. Can we do anything from this? Is there any elegant and GOOD solution for this question? I failed to do anything but spending three hours.

Please help me thank you

elementary-number-theory polynomials

share|cite|improve this question

edited Mar 21 at 5:14

Sufaid Saleel

asked Mar 21 at 5:09

Sufaid SaleelSufaid Saleel

1,756829

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Technically, the expressions you have need to be four times some perfect square, some integers $m,n$ such that $(a^2 - 4b) = 4m^2, (4a^2 - 4b) = 4n^2$. That way when you take the square root, you have $2n,2m$. The quadratic formula requires you to divide by two and you want your roots to be integers. (Also the latter expression, $(4a^2 - 4b)$, can have its four factored out, so you just really need $(a^2 - b) = n^2$ for that as a result. ... Not that this is a ton of help but it's worth noting nonetheless
  $endgroup$
  – Eevee Trainer
  Mar 21 at 5:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @EeveeTrainer The division by $2$ isn't a problem - because we also have that linear term $-a$ before we divide by $2$. If $sqrta^2-4b$ is an odd integer, $-a+sqrta^2-4b$ and $-a-sqrta^2-4b$ are both even.
  $endgroup$
  – jmerry
  Mar 21 at 5:31
  

  
  
  
  

add a comment | 

$begingroup$

Show that there are infinitely many pairs $(a,b)$ of relatively prime integers (not necessarily positive) such that both quadratic equations $x^2+ax+b=0$ and $x^2+2ax+b=0$ have integer roots.

I noticed the obvious relation that $(a^2-4b)$ and $(4a^2-4b)$ both are perfect squares. Can we do anything from this? Is there any elegant and GOOD solution for this question? I failed to do anything but spending three hours.

Please help me thank you

elementary-number-theory polynomials

share|cite|improve this question

edited Mar 21 at 5:14

Sufaid Saleel

asked Mar 21 at 5:09

Sufaid SaleelSufaid Saleel

1,756829

$endgroup$

share|cite|improve this question

edited Mar 21 at 5:14

Sufaid Saleel

asked Mar 21 at 5:09

Sufaid SaleelSufaid Saleel

1,756829

share|cite|improve this question

edited Mar 21 at 5:14

Sufaid Saleel

asked Mar 21 at 5:09

Sufaid SaleelSufaid Saleel

1,756829

asked Mar 21 at 5:09

Sufaid SaleelSufaid Saleel

1,756829

asked Mar 21 at 5:09

Sufaid SaleelSufaid Saleel

1,756829

1 Answer
1

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oldest

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2

$begingroup$

We want $a^2-4b$ and $4a^2-4b$ to both be perfect squares $m^2$ and $n^2$. So then, their difference is $3a^2$, and we know how to factor a perfect square: $n^2-m^2=(n+m)(n-m)$. Can we assign those to two factors of $3a^2$?

First, we try the pair $a$ and $3a$. Then $n+m=3a$ and $n-m=a$, so $n=2a$ and $m=a$. This gives $b=0$. Yes, the pairs $(a,0)$ and $(2a,0)$ lead to integer solutions - but they're not relatively prime. We'll need relatively prime factors to make things work.

Next, the pair $a^2$ and $3$. Then (assuming $|a|>1$), $n+m=a^2$ and $n-m=3$, so $n=fraca^2+32$ and $m=fraca^2-32$. We need $a$ odd for this to be an integer, but it looks good so far. From $a^2-4b=m^2=fraca^4-6a^2+94$, we get $b=frac-a^4+10a^2-916=frac-(a^2-1)(a^2-9)16$. If $a$ is divisible by $3$, that will have a common factor $3$ with $a$; otherwise, it will be relatively prime.

So there it is: one infinite family. Choose $a$ to be an integer not divisible by $2$ or $3$, and set $b=frac-(a^2-1)(a^2-9)16$. The roots of $x^2+ax+b=0$ are $frac(a-3)(a+1)4$ and $frac-(a+3)(a-1)4$, while the roots of $x^2+2ax+b=0$ are $frac(a-3)(a-1)4$ and $frac-(a+3)(a+1)4$. A table of the first few:

$$beginarrayca & b & textRoots of x^2+ax+b=0 & textRoots of x^2+2ax+b=0\
hline 1 & 0 & -1,0 & 0,-2\ 5 & 24 & 3,-8 & 2,-12 \ 7 & 120 & 8,-15 & 6,-20 \ 11 & 840 & 24,-35 & 20,-42 \ 13 & 1680 & 35,-48 & 30,-56 \ 17 & 5040 & 63,-80 & 56,-90 \ 19 & 7920 & 80,-99 & 72,-110 endarray$$
Are there other solutions? Of course there are - but we don't need to find them. We have enough already.

share|cite|improve this answer

answered Mar 21 at 6:08

jmerryjmerry

16.9k11633

$endgroup$

add a comment | 

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2

$begingroup$

We want $a^2-4b$ and $4a^2-4b$ to both be perfect squares $m^2$ and $n^2$. So then, their difference is $3a^2$, and we know how to factor a perfect square: $n^2-m^2=(n+m)(n-m)$. Can we assign those to two factors of $3a^2$?

First, we try the pair $a$ and $3a$. Then $n+m=3a$ and $n-m=a$, so $n=2a$ and $m=a$. This gives $b=0$. Yes, the pairs $(a,0)$ and $(2a,0)$ lead to integer solutions - but they're not relatively prime. We'll need relatively prime factors to make things work.

Next, the pair $a^2$ and $3$. Then (assuming $|a|>1$), $n+m=a^2$ and $n-m=3$, so $n=fraca^2+32$ and $m=fraca^2-32$. We need $a$ odd for this to be an integer, but it looks good so far. From $a^2-4b=m^2=fraca^4-6a^2+94$, we get $b=frac-a^4+10a^2-916=frac-(a^2-1)(a^2-9)16$. If $a$ is divisible by $3$, that will have a common factor $3$ with $a$; otherwise, it will be relatively prime.

So there it is: one infinite family. Choose $a$ to be an integer not divisible by $2$ or $3$, and set $b=frac-(a^2-1)(a^2-9)16$. The roots of $x^2+ax+b=0$ are $frac(a-3)(a+1)4$ and $frac-(a+3)(a-1)4$, while the roots of $x^2+2ax+b=0$ are $frac(a-3)(a-1)4$ and $frac-(a+3)(a+1)4$. A table of the first few:

$$beginarrayca & b & textRoots of x^2+ax+b=0 & textRoots of x^2+2ax+b=0\
hline 1 & 0 & -1,0 & 0,-2\ 5 & 24 & 3,-8 & 2,-12 \ 7 & 120 & 8,-15 & 6,-20 \ 11 & 840 & 24,-35 & 20,-42 \ 13 & 1680 & 35,-48 & 30,-56 \ 17 & 5040 & 63,-80 & 56,-90 \ 19 & 7920 & 80,-99 & 72,-110 endarray$$
Are there other solutions? Of course there are - but we don't need to find them. We have enough already.

share|cite|improve this answer

answered Mar 21 at 6:08

jmerryjmerry

16.9k11633

$endgroup$

add a comment | 

2

$begingroup$

We want $a^2-4b$ and $4a^2-4b$ to both be perfect squares $m^2$ and $n^2$. So then, their difference is $3a^2$, and we know how to factor a perfect square: $n^2-m^2=(n+m)(n-m)$. Can we assign those to two factors of $3a^2$?

First, we try the pair $a$ and $3a$. Then $n+m=3a$ and $n-m=a$, so $n=2a$ and $m=a$. This gives $b=0$. Yes, the pairs $(a,0)$ and $(2a,0)$ lead to integer solutions - but they're not relatively prime. We'll need relatively prime factors to make things work.

Next, the pair $a^2$ and $3$. Then (assuming $|a|>1$), $n+m=a^2$ and $n-m=3$, so $n=fraca^2+32$ and $m=fraca^2-32$. We need $a$ odd for this to be an integer, but it looks good so far. From $a^2-4b=m^2=fraca^4-6a^2+94$, we get $b=frac-a^4+10a^2-916=frac-(a^2-1)(a^2-9)16$. If $a$ is divisible by $3$, that will have a common factor $3$ with $a$; otherwise, it will be relatively prime.

So there it is: one infinite family. Choose $a$ to be an integer not divisible by $2$ or $3$, and set $b=frac-(a^2-1)(a^2-9)16$. The roots of $x^2+ax+b=0$ are $frac(a-3)(a+1)4$ and $frac-(a+3)(a-1)4$, while the roots of $x^2+2ax+b=0$ are $frac(a-3)(a-1)4$ and $frac-(a+3)(a+1)4$. A table of the first few:

$$beginarrayca & b & textRoots of x^2+ax+b=0 & textRoots of x^2+2ax+b=0\
hline 1 & 0 & -1,0 & 0,-2\ 5 & 24 & 3,-8 & 2,-12 \ 7 & 120 & 8,-15 & 6,-20 \ 11 & 840 & 24,-35 & 20,-42 \ 13 & 1680 & 35,-48 & 30,-56 \ 17 & 5040 & 63,-80 & 56,-90 \ 19 & 7920 & 80,-99 & 72,-110 endarray$$
Are there other solutions? Of course there are - but we don't need to find them. We have enough already.

share|cite|improve this answer

answered Mar 21 at 6:08

jmerryjmerry

16.9k11633

$endgroup$

add a comment | 

$begingroup$

We want $a^2-4b$ and $4a^2-4b$ to both be perfect squares $m^2$ and $n^2$. So then, their difference is $3a^2$, and we know how to factor a perfect square: $n^2-m^2=(n+m)(n-m)$. Can we assign those to two factors of $3a^2$?

First, we try the pair $a$ and $3a$. Then $n+m=3a$ and $n-m=a$, so $n=2a$ and $m=a$. This gives $b=0$. Yes, the pairs $(a,0)$ and $(2a,0)$ lead to integer solutions - but they're not relatively prime. We'll need relatively prime factors to make things work.

Next, the pair $a^2$ and $3$. Then (assuming $|a|>1$), $n+m=a^2$ and $n-m=3$, so $n=fraca^2+32$ and $m=fraca^2-32$. We need $a$ odd for this to be an integer, but it looks good so far. From $a^2-4b=m^2=fraca^4-6a^2+94$, we get $b=frac-a^4+10a^2-916=frac-(a^2-1)(a^2-9)16$. If $a$ is divisible by $3$, that will have a common factor $3$ with $a$; otherwise, it will be relatively prime.

So there it is: one infinite family. Choose $a$ to be an integer not divisible by $2$ or $3$, and set $b=frac-(a^2-1)(a^2-9)16$. The roots of $x^2+ax+b=0$ are $frac(a-3)(a+1)4$ and $frac-(a+3)(a-1)4$, while the roots of $x^2+2ax+b=0$ are $frac(a-3)(a-1)4$ and $frac-(a+3)(a+1)4$. A table of the first few:

$$beginarrayca & b & textRoots of x^2+ax+b=0 & textRoots of x^2+2ax+b=0\
hline 1 & 0 & -1,0 & 0,-2\ 5 & 24 & 3,-8 & 2,-12 \ 7 & 120 & 8,-15 & 6,-20 \ 11 & 840 & 24,-35 & 20,-42 \ 13 & 1680 & 35,-48 & 30,-56 \ 17 & 5040 & 63,-80 & 56,-90 \ 19 & 7920 & 80,-99 & 72,-110 endarray$$
Are there other solutions? Of course there are - but we don't need to find them. We have enough already.

share|cite|improve this answer

answered Mar 21 at 6:08

jmerryjmerry

16.9k11633

$endgroup$

share|cite|improve this answer

answered Mar 21 at 6:08

jmerryjmerry

16.9k11633

answered Mar 21 at 6:08

jmerryjmerry

16.9k11633

answered Mar 21 at 6:08

jmerryjmerry

16.9k11633