Characterizing integer solutions to $a^2 \mid \bigl((b^2)^2 + (b^2+1)^2\bigr)$

Question

I’m looking to characterize all integers $a$ and $b$ satisfying $$a^2 \mid \bigl((b^2)^2 + (b^2+1)^2\bigr), \qquad a > b \ge 1.$$

Brute force searches have so far turned up the two solutions $(a,b) = (13,7)$ and $(a,b)=(9601,5441)$. I’m guessing there are an infinite number, but I’d like to reduce the numbers I have to check. Obviously, $a$ must be odd and the sum of two squares, so I’m already taking that into account.

Any thoughts about what the next solution might be, or how to attack the characterization of any/all solutions?

If $b$ is odd, then there exists an integer $c$ such that $a=b+2c$. Substituting, we find $a \parallel (32c^4+8c^2+1) = \bigl((4c^2)^2+(4c^2+1)^2\bigr)$, and then $a^2$ divides a related quotient. Based on that, I feel like there might be a Vieta-jumping solution… but I still can’t quite find it.

Answer 1

When there is very little output in a raw search, it becomes worthwhile to weaken restrictions, see what can be learned from a larger collection of second-best answers. Below, the first number is $b,$ the next is a square $d>b$ that divides $2b^4 + 2b^2 + 1.$ When the stronger condition $d>b^2$ also holds...

7  169 =   13^2    SWING LOW SWEET CHARIOT  
11  25 =   5^2   
14  25 =   5^2   
65  841 =   29^2   
162  169 =   13^2   
162  1681 =   41^2   
162  284089 =   13^2 41^2    SWING LOW SWEET CHARIOT  
186  625 =   5^4   
270  3721 =   61^2   
272  1369 =   37^2   
363  1681 =   41^2   
367  18769 =   137^2   
439  625 =   5^4   
514  4225 =   5^2 13^2   
776  841 =   29^2   
914  100489 =   317^2   
914  2512225 =   5^2 317^2    SWING LOW SWEET CHARIOT  
946  2809 =   53^2   
1097  1369 =   37^2   
1318  1681 =   41^2   
1514  4225 =   5^2 13^2   
1519  1681 =   41^2   
1835  167281 =   409^2   
1863  2809 =   53^2   

Answer 2

This problem in a way is related to equation $x^2+(x+1)^2=y^2$ which has infinite solution.There is a proof in this site.I will try to show this relation by an examples:

$119^2+120^2=13^4$

$119=7\times 17$

$(7\times17)^2+(7\times17+1)=13^4$$\space\space\space\space\space\space(1)$

and we see that:

$7^4+(7^2+1)^2=29\times 13$

Relation (1) is symmetric for 7 and 17 so relation $b^4+(b^2+1)^2= km^2$ must also be true for 17, we check this:

$17^4+(17^2+1)^2=86021=509\times 13^2$

that is if $b$ is a primes multiple of a primitive solution, that prime is also a solution. $x$ and $x+1$ make following sets:

$x\in\{3, 119, 4059, 137903\cdot\cdot\cdot\}$

$(x+1)\in\{4, 120, 4060, 137904\cdot\cdot\cdot\}$

Answer 3

$a^2\cdot n = 2b^4 + 2b^2 + 1.$

Factorization method:

After prime factorization of $2b^4 + 2b^2 + 1,$ divide them into two groups $(a,n).$
If the power of prime number $\gt 1$ then it is included in group $a.$

Example:
Take $b=131972,$ then $2b^4 + 2b^2 + 1 = 29^2\cdot 11057\cdot158621^2\cdot2593.$
Thus one solution is $a=158621, n=24112143641.$
As Will Jagy mentions this solution, it has another solution $(a,n)=(29\cdot 158621=4600009,28670801).$

           [a,b,n],b<500000
           
           [13, 7, 29]
           [533, 162, 4849]
           [1585, 914, 555593]
           [9601, 5441, 19015685]
           [14045, 6564, 18821785]
           [158621, 131972, 24112143641]
           [218233, 114689, 7265679125]
           [1091165, 114689, 290627165]
           [4600009, 131972, 28670801]

Answer 4

We know that $(b)^2$ and $(b+1)^2$ are of opposite parity and this makes the sum of their squares odd so $(a)^2$ must be odd. Brute force, even limited to $15$ digits, shows solutions not in the OP question.

Here are the first $3$ $\space (a,b)$-values where $\quad 2 b^4 + 2 b^2 + 1 \le 987654321054321.$ $$ (13 ,7 )\quad (533 ,162 )\quad (1585 ,914 )\quad $$ A sum need not be a perfect square but only contain a perfect square as one of its factors. Here: $\quad f(7)/13^2=29\\ f(162)/533^2=4849\\ f(914)/1585^2=555593$

and here is the program that generated them

  10 print "Enter H1" : input h1 : print
  20 for a1 = 3 to h1
  30    for b1 = 2 to a1-1
  40    if 2*b1^4+2*b1^2+1 <= 987654321054321
  50       d1 = (2*b1^4+2*b1^2+1)/a1^2
  60       if d1 = int(d1)
  70          print "(" a1 "," b1 ")\quad "
  80       endif
  90     endif
  100    next b1
  110 next a1

Note: In an earlier edit of this answer I overlooked rounding errors that made most of my solutions invalid where $\quad 2 b^4 + 2 b^2 + 1\ge 10^{15},\quad $ I have limited my "solutions" here but, in any language with arbitrary precision, this restriction should not be needed.

Answer 5

If $a^2 | 2b^4+2b^2+1$, then for any prime factor $p$ of $a$:

$ p | 2b^4+2b^2+1$

We can find the possible values of $b$ modulo $p$. Values only occur when $p = 1$ mod 4, and there are either 0,2, or 4 possible values, as shown in the following analysis:

Set $ X = b^2$, then $2X^2+2X+1 = 0$ mod $p$

Multiply by 2 (clearly p is odd) and rearrange to get $(2X+1)^2 = -1$ mod $p$

Find $w$, a fourth root of unity mod $p$ (this is only possible if $p = 1$ mod 4), then

$2X+1 = \pm w$ mod $p$

rearranging, $b^2 = X = (\pm w-1)(\frac{p+1}{2})$ mod $p$

$b$ has 0, 2 or 4 solutions mod p, depending on whether there are solutions to the above equation.

This suggests an efficient search algorithm:

1. for primes $p = 1$ mod 4, find and store the 2 or 4 values of $b$ mod $p$

2. for $b$ in a range of values, find the primes $p$ that satisfy $ p | 2b^4+2b^2+1$,

3. from those, find the ones where $ p^{2n} | 2b^4+2b^2+1$

4. using the values $p,b$ found in step 3, construct values $a$ (primes or composite) satisfying $ a^2 | 2b^4+2b^2+1 $

I have implemented this a python 3 program, searching primes up to $10^8$ and values of $b$ up to $10^8$. The only solutions found have been those already identified by @Tomita

The Python 3 program is on github https://github.com/armchaircaver/Characterizing-integer-solutions-to-a-squared-divides-polynomial-in-b

Addition 30 Aug 2021:

There are techniques to find a square root modulo a prime power (see for example https://www.johndcook.com/blog/quadratic_congruences/ ), which i have used to find values of b modulo $p^2$ satisfying $p^2 | 2b^4+2b^2+1$ directly, making the search even more efficient, as there are 2 or 4 values of $b$ mod $p^2$. The more efficient version is "stackexchange puzzle sieve p squared.py", and has been added to the same github repository as above.

Using the more efficient version I have searched primes up to $10^8$ and values of $b$ up to $10^9$ with no new solutions found.