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Based on experimental data for primes $< 1.4 \times 10^{10}$, I observed that

Every natural number $x$ which is a solution of $x^2 = y^2 - z^2$ where $y$ and $z$ are primes $> 5000$ has a prime factor greater than $17$.

Is this true in general or can we have a counter example?

Note: Posted in MO since it is unanswered in MSE

Code: Generates all all solutions in which the largest prime factor of $x$ is less than $101$. 

s = 5
i = 1
f = 1
target = set = 10^6
q_max = 0
while True:
    if s*(s+1)%30 == 0:
        q = 2*s + 1
        p = 2*s^2 + q
        n = p - 1       
        if is_prime(p) and is_prime(q):
            i = i + 1
            F = prime_divisors(n)
            if F[-1] <= 101:
                f = f + 1
                q_max = q
                print (i,s,f,n,p,q_max, F[-1])
    if s > target:
        print "Reached", target, f,q_max
        target = target + set
    s = s + 1
$\endgroup$
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  • 1
    $\begingroup$ what does $x^2=(y-z)(y+z)$ tell you $\endgroup$
    – user645636
    Dec 20, 2019 at 13:01
  • 5
    $\begingroup$ @RoddyMacPhee; Well, what does it tell you? $\endgroup$
    – TonyK
    Dec 20, 2019 at 13:13
  • 6
    $\begingroup$ As a vague argument in support of this sort of conclusion: We note that $x^2+z^2=y^2$ so we have a Pythagorean triple. Hence we can find integers $m,n$ such that $$x=2mn\quad y=m^2+n^2\quad z=m^2-n^2$$ Where we have used the fact that the conditions imply that $x$ is even. Since $z=(m+n)(m-n)$ is prime we must have $m=n+1$. Thus $$x=2n(n+1)\quad y=2n^2+2n+1\quad z=2n+1$$ In particular, $x$ is divisible by the product of two consecutive integers, $n, n+1$. It is very difficult for two, large, consecutive numbers to both be divisible only by small primes. $\endgroup$
    – lulu
    Dec 20, 2019 at 13:15
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    $\begingroup$ it tells you that $z$ has to stay clear of two modular classes ( related to the class of $y$) for every possible prime $>17$ or that it only appears in exactly one factor ,once with multiplicity... @TonyK $\endgroup$
    – user645636
    Dec 20, 2019 at 13:20
  • 2
    $\begingroup$ @RoddyMacPhee I don't see how you arguments leads to the proof but in case you have a complete proof, why don't you post it $\endgroup$
    – Nilotpal Sinha
    Dec 21, 2019 at 2:59

1 Answer 1

Reset to default
5
$\begingroup$

As stated in lulu's comment, we know that $x,y,z$ are a Pythagorean triple and that we require $$x=2mn\quad y=m^2+n^2\quad z=m^2-n^2\quad m=n+1$$

By Størmer's theorem and https://oeis.org/A117581 we know that the largest consecutive $17$-smooth integers are $336140$ and $336141$. Testing all pairs of consecutive $17$-smooth integers up to this limit produces the largest answer $$12495000^2=12495001^2-4999^2$$ so the original observation is true.

Raw output from program for consecutive $17$-smooth integers $\ge2499$ :

n    m       x        y        z     Prime y? Prime z?
2499 2500    12495000 12495001 4999    True True
2600 2601    13525200 13525201 5201    False False
3024 3025    18295200 18295201 6049    False False
4095 4096    33546240 33546241 8191    False True
4224 4225    35692800 35692801 8449    False False
4374 4375    38272500 38272501 8749    True False
4913 4914    48284964 48284965 9827    False False
5831 5832    68012784 68012785 11663    False False
6655 6656    88591360 88591361 13311    False False
9800 9801    192099600 192099601 19601    False False
10647 10648    226738512 226738513 21295    False False
12375 12376    306306000 306306001 24751    True False
14399 14400    414691200 414691201 28799    False False
28560 28561    1631404320 1631404321 57121    False False
31212 31213    1948440312 1948440313 62425    False False
37179 37180    2764630440 2764630441 74359    False False
123200 123201    30356726400 30356726401 246401    False False
194480 194481    75645329760 75645329761 388961    False True
336140 336141    225980871480 225980871481 672281    False False
$\endgroup$

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