$a^2 = 3^b + 37$

Question

Find all solutions in integers to $a^2 = 3^b + 37$.

I came up with this problem myself. (I chose $37$ so there would be a relatively large solution.) But is there a nice way to approach this problem? I got that $a$ must be divisible by $4$ and that $b$ must be odd, but I couldn't progress past this. I tried making a substitution out of this, saying that $a = 4x$ and $b = 2y + 1$ but I couldn't quite get this to work.

For reference, Wolfram Alpha says there are four solutions: (a, b) = (8, 3); (-8, 3); (3788, 15); (-3788, 15)

Working backwards from the solutions, it would appear that $b\equiv 3\mod 12$ is a starting point. What sort of answer are you hoping for, knowing that WA seems to suggest there are only four solutions? — abiessu, Dec 05 '24 at 04:23
FYI, using an Approach0 search, there's the fairly similar Finding positive integer solutions to $3^x + 55=y^2$, but the exponent there must be even, so the solution involves a product of the difference of squares equaling $55$, so it's considerably simpler. In some ways, a more similar type question to yours is Perhaps a Pell type equation, where the exponent is odd like in your case. Solving it as a Pell's equation is suggested, but ... — John Omielan, Dec 05 '24 at 04:24
(cont.) the one solution there instead uses elliptic curves, or Mordell equations. — John Omielan, Dec 05 '24 at 04:25
Let $y=3^m$ & $b=3m+r$, where r=0,1,2 it reduces to $3^r ×y^m +37=a^2$ is an elliptic curve. You need to find all Integer points for this elliptic curve. By Siegel theorem: If an Elliptic curve has a nonzero discriminant then Number of integer points of an Elliptic Curve is finite. — Guruprasad, Dec 05 '24 at 13:28
These kind of equations are called Ramanujan - Nagell type Equations — Guruprasad, Dec 05 '24 at 13:34
There is a typo in my 1st comment in 3rd line. It should be :$a^2=3^r × y^3+37$ — Guruprasad, Dec 05 '24 at 18:39

score 0 · Answer 1 · answered Dec 05 '24 at 04:34

from decimal import Decimal, getcontext

def find_values():
    results = []
    getcontext().prec = 1000  # Set precision high enough
    for b in range(1, 1000, 2):  # b must be odd, so we step by 2
        value = Decimal(3)**b + 37
        a = int(value.sqrt())
        if a % 4 == 0 and a**2 == value:
            results.append((a, b))
    return results

values = find_values()
for a, b in values:
    print(f"a = {a}, b = {b}")

if not values:
    print("No values found.")

a = 8, b = 3
a = 3788, b = 15

I think that [a = 8, b = 3] , [a = 3788, b = 15] (and it's negative values for a) are only that work for that sequence

code only gives positive values !

score 0 · Answer 2 · answered Dec 05 '24 at 09:55

Comment: Starting from the solutions you found we may try to pridict the existence of more solutions. We may manipulate like this:

$3^b+3^3=a^2-10\Rightarrow a^2-10\geq 27$

In the case $a^2=27$ we have:

$27=3^b\Rightarrow b=3$

Which gives $(a, b)=(8, 3)$

We may also say $3\mid a^2-37$ and a is even. For searching we write:

$a=8, 10, 12, 14, 16\cdot\cdot 28\cdot\cdot\cdot$

$a^2= 64, 100, 144, 256, \cdot\cdot\cdot784$

then try to find a sequence or a progression, for example take $64$ as the first number of a progression:

$100=64+4\times 9$

$196=54+4\times 33$

$256=64+4\times 48$

$\cdot$

$724=64+4\times 165$

$\cdot$

$3788^2=64+ 4\times 3587220$

Now in the progression formula $N= a+n \cdot d$ we have $a=64$ and $d=4$ and we have to find a relation between n and N. $n$ is not the number of term or elements but it has the following sequence:

$9, 33, 48, 84, 108, 153, 165, \cdot\cdot\cdot 3587220\cdot\cdot\cdot \infty $

The number $3587220$ gives $a^2= 64+4\times 3587220=3788^2$ which gives $(a, b)(3788, 15)=15$.

That means more solutions are probable as Kodzis found in his second post.

$a^2 = 3^b + 37$

2 Answers2