Problems & Puzzles: Puzzles

Puzzle 1113 A pair of Diophantine equations.

Paul Cleary sent the following nice puzzle:
 

Given the two equations

 

x^2 + 1200*y = n^2

y^2 + 1200*x = m^2

 

x > y, and both positive integers.

 

When and x = 157 and y = 143 both n and m are the primes 443 and 457 respectively.

 

Q1. Find more values for x and y, where x > y that make both n and m primes.

Q2. Can you tell how many distinct prime-solutions in total are there for this pair of equations?


During the week Nov 26-Dic 2, 2022, contributions came from Giorgos Kalogeropoulos, Paul Cleary, Adam Stinchcombe, Gennady Gusev, Ken Wilke, Emmanuel Vantieghem, Oscar Volpatti

***

Giorgos wrote:

These two families of curves (where the second one is rotated 90 degrees) intersect infinitely often.
This is why I believe we can find infinitely many lattice points {x,y} although they are pretty sparse.
Here are the 41 solutions that I found with n and m primes:

 
{  x  ,  y  ,  n  ,  m }
{157,143,443,457}
{161,139,439,461}
{167,133,433,467}
{179,121,421,479}
{191,109,409,491}   x,y are also primes
{199,101,401,499}  x,y are also primes
{203,97,397,503}
{221,79,379,521}
{241,59,359,541}   x,y are also primes
{247,53,353,547}
{263,37,13,313}   x,y are also primes
{269,31,331,569}  x,y are also primes
{287,13,313,587}
{293,7,307,593}   x,y are also primes
{407,169,607,719}
{1469,523,1669,1427}
{3689,2743,4111,3457}
{4301,2611,4651,3461}
{5269,2821,5581,3779}
{5819,487,5869,2687}
{6649,5773,7151,6427}
{10421,139,10429,3539}
{15181,2233,15269,4817}
{19409,19109,19991,19709}
{27071,18547,27479,19403}
{53983,16583,54167,18433}
{96889,3553,96911,11353}
{108521,1447,108529,11503}
{160867,17699,160933,22501}
{170051,14173,170101,20123}
{171127,98783,171473,99817}
{212471,2833,212479,16217}
{261299,871,261301,17729}
{1085171,90433,1085221,97367}
{1724659,235709,1724741,240059}
{1906159,737093,1906391,738643}
{1926173,815467,1926427,816883}
{2156089,79057,2156111,94007}
{2991173,518479,2991277,521929}
{25257869,2609983,25257931,2615783}
{113476571,1513021,113476579,1557371}

***

Paul wrote:

This is my method and solutions to this puzzle.

 

Given

 

x^2 + 1200y = n^2

y^2 + 1200x = m^2

 

since x and y are positive we can write

 

x^2 + 1200y = (x + a)^2 and

y^2 + 1200x = (y + b)^2.

 

Where a and b are positive integers, after expanding we find the square terms cancel leaving just linear simultaneous equations thus.

 

 

1200y = 2 a x + a^2

1200x = 2 b y + b^2

 

Solving for x and y we get the following

 

x = (2*a^2 b + 1200 b^2)/(1440000 - 4 a b) and

y = (2*a b^2 + 1200 a^2)/(1440000 - 4 a b).

 

and since a and b are positive, the numerators in each fraction must be positive, so to have a positive denominator

a*b < 360000.  with Judicial selection of a and b, all solutions (3535 in total) can be found in under a minute.  However of those 3535 solutions only 48 with x>y>0 and both m and n prime.

 

Here are all my solutions.

 

...

Regarding the solutions sent by Giorgos, Paul noted:

All the solutions submitted are correct, though there are 7 solutions missing Im, surprised Giorgos missed this one.  {313, 13, 337, 613}. Then there are these after his last solution.

 

{210972899, 703243, 210972901, 864707}

{213602921, 2848039, 213602929, 2892689}

{271181099, 903937, 271181101, 1068887}

{995465099, 3318217, 995465101, 3493583}

{7287489299, 24291631, 7287489301, 24470969}

{24246030899, 80820103, 24246030901, 80999903}.

 

Regarding the claim of infinite solutions, or lattice points for {x,y} I dont think there are, there are only some 3535 solutions in total, If there are I would be interested in any with an x, y value greater than, x = 2429946090299 and y = 8099820301.

***

Adam wrote:

Up through max(n,m)=the 1807 th prime, I found 24 sets of n,m,x,y:
 
[443, 457, 157, 143], [439, 461, 161, 139], [433, 467, 167, 133], [421, 479, 179, 121], [409, 491, 191, 109], [401, 499, 199, 101], [397, 503, 203, 97], [379, 521, 221, 79], [359, 541, 241, 59], [353, 547, 247, 53], [337, 563, 263, 37], [331, 569, 269, 31], [313, 587, 287, 13], [307, 593, 293, 7], [337, 613, 313, 13], [607, 719, 407, 169], [1427, 1669, 523, 1469], [3457, 4111, 2743, 3689], [3461, 4651, 2611, 4301], [3779, 5581, 2821, 5269], [2687, 5869, 487, 5819], [6427, 7151, 5773, 6649], [3539, 10429, 139, 10421], [4817, 15269, 2233, 15181]
 

 
For seven of these, all x,y,n,m are prime.

***

Gennady wrote:

Q1. There were found 34 solutions for x,y < 10^6.
    x       y       n       m
------  ------  ------  ------
   157     143     443     457
   161     139     439     461
   167     133     433     467
   179     121     421     479
   191     109     409     491
   199     101     401     499
   203      97     397     503
   221      79     379     521
   241      59     359     541
   247      53     353     547
   263      37     337     563
   269      31     331     569
   287      13     313     587
   293       7     307     593
   313      13     337     613
   407     169     607     719
  1469     523    1669    1427
  3689    2743    4111    3457
  4301    2611    4651    3461
  5269    2821    5581    3779
  5819     487    5869    2687
  6649    5773    7151    6427
 10421     139   10429    3539
 15181    2233   15269    4817
 19409   19109   19991   19709
 27071   18547   27479   19403
 53983   16583   54167   18433
 96889    3553   96911   11353
108521    1447  108529   11503
160867   17699  160933   22501
170051   14173  170101   20123
171127   98783  171473   99817
212471    2833  212479   16217
261299     871  261301   17729
 

***

Ken wrote:

Attached is my comment on Puzzle 1113. It finds other members of a family of solutions based upon the example solution given. Here may be additional soluti9ons, but I did not see any decent route to them.

Solution:
In this solution, we consider only those solutions which are the same type as the example
given by the proposer. Consider the following equations in which x, y, m and n and all
positive integers with x &gt;y, m&gt;n and both m and n are primes.
x^2 + 1200*y = n^2 (1)
y^2 + 1200*x = m^2 (2)
Subtracting equation (2) from equation (1) yields
x^2y^2 + 1200*(y x) = n^2 m^2 which can be rewritten as
(xy)[(x+y)-1200] = (n-m)(m+n) (3)
According to the sample solution provided, we have x-y = m-n. Thus since m+n &gt; x+y,
x+ y &gt; 1200 is impossible so equation (3) becomes
1200 = (m+n)+(x+y) (4).
Noting that in the given solution, m+n =457 + 443 = 900 and x+y = 300 we search a table
of primes for pairs of primes m and n such that m+n =900. That search reveals 14 pairs of
primes, including the pair in the given solution,(m,n) =(457,443), (461,439), (467,433),
(479, 421), (491,409),(499, 401), (503,397), (521,379), (541,359), (547,353), (563,337),
(569,331), (587,313)and (593, 307). The corresponding pairs of x and y are respectively
(x,y) can be computed easily since m-n = x-y and m+n =900.
Then (m,n,n,x,y)=(457,443,157,143),(461,439,161,139),(467,433,161,139),
(479,421,179,121),((491,409,191,109),(499,401,199,101),(503,397,203,97),
(521,379,221,79),(541,359,241,59),(547,353,247,53),(563,337,263,37),(569,331,269,31),
(587,313,287,13)and (593,307,293,7).

Of particular interest are (m,n,x,y) =
(491,409,191,109),(499,401,199,101),(503,397,203,97),(541,359,241,59),
(563,337,263,37),(569,331,269,31),(587,313,287,13) and (593,307,293,7)
in which each of m, n, x and y are prime.

***

Emmanuel wrote:

Q1
Here are all the solution I found for  x, y, m, n :
157, 143, 457, 443
161, 139, 461, 439
167, 133, 467, 433
179, 121, 479, 421
191, 109, 491, 409 *
199, 101, 499, 401 *
203, 97, 503, 397
221, 79, 521, 379
241, 59, 541, 359 *
247, 53, 547, 353
263, 37, 563, 337 *
269, 31, 569, 331 *
287, 13, 587, 313
293, 7, 593, 307 *
313, 13, 613, 337 *
407, 169, 719, 607
1469, 523, 1427, 1669
3689, 2743, 3457, 4111
4301, 2611, 3461, 4651
5269, 2821, 3779, 5581
5819, 487, 2687, 5869
6649, 5773, 6427, 7151
10421, 139, 3539, 10429
15181, 2233, 4817, 15269
19409, 19109, 19709, 19991
27071, 18547, 19403, 27479
53983, 16583, 18433, 54167
96889, 3553, 11353, 96911
108521, 1447, 11503, 108529
160867, 17699, 22501, 160933
170051, 14173, 20123, 170101
171127, 98783, 99817, 171473
212471, 2833, 16217, 212479
261299, 871, 17729, 261301
1085171, 90433, 97367, 1085221
1724659, 235709, 240059, 1724741
1906159, 737093, 738643, 1906391
1926173, 815467, 816883, 1926427
2156089, 79057, 94007, 2156111
2991173, 518479, 521929, 2991277
25257869, 2609983, 2615783, 25257931
113476571, 1513021, 1557371, 113476579
210972899, 703243, 864707, 210972901
213602921, 2848039, 2892689, 213602929
271181099, 903937, 1068887, 271181101
(The seven solutions marked with  *  consist of four primes).
 

 
Q2.
I think there might be infinitely many solutions.

***

Oscar wrote:

There are only 48 pairs (x,y) with 0<y<x such that n and m are both prime.
Let's assume that positive integers x,y,n,m are a solution. 
n^2 = x^2+1200*y > x^2,
m^2 = y^2+1200*x > y^2.
n is greater than x and has the same parity of x, so n = x+2*a for some positive integer a;
m is greater than y and has the same parity of y, so m = y+2*b for some positive integer b.

By substituting and simplifying, we obtain a system of equations which is linear in x and y:
300*y = a*x + a^2,
300*x = b*y + b^2. 
If a*b = 90000 the system is inconsistent, otherwise it has a unique solution: 
x = b*(300*b+a^2)/(90000-a*b),
y = a*(300*a+b^2)/(90000-a*b).


I'll prove that the upper bound a*b < 90000 must hold.

Consider the first equation:
300*y = a*x + a^2.

Solve for x in terms of y:
x = (300/a)*y - a.
Drop the term "- a", obtaining a strict inequality:
x < (300/a)*y.
Now consider the second equation:
(y+2*b)^2 = y^2 + 1200*x.
Substitute "x" with "(300/a)*y", obtaining another strict inequality:  
(y+2*b)^2 < y^2 + (360000/a)*y.

Add the positive term "(180000/a)^2" to the right side only, so that inequality still holds:
(y+2*b)^2 < y^2 + (360000/a)*y + (180000/a)^2.

Now the right side is the square of a positive quantity too:
(y+2*b)^2 < (y+180000/a)^2.

Take the square root of both sides, then cancel y out:
2*b < 180000/a.

Multiply both sides by "a/2":
a*b < 90000.
QED.


Therefore, only 1040555 pairs (a,b) are permitted.
For each permitted pair of positive integers (a,b), we obtain a unique pair of positive rationals (x,y).
Inequality y<x also holds if and only if a<b; this further reduces the search space to 520128 pairs (a,b).
Only 3535 permitted pairs (a,b) actually produce an integer pair (x,y).

Only 48 such pairs (a,b) also produce a prime pair (n,m).


x  y  n  m
157  143  443  457
161  139  439  461
167  133  433  467
179  121  421  479
191  109  409  491
199  101  401  499
203  97  397  503
221  79  379  521
241  59  359  541
247  53  353  547
263  37  337  563
269  31  331  569
287  13  313  587
293  7  307  593
313  13  337  613
407  169  607  719
1469  523  1669  1427
3689  2743  4111  3457
4301  2611  4651  3461
5269  2821  5581  3779
5819  487  5869  2687
6649  5773  7151  6427
10421  139  10429  3539
15181  2233  15269  4817
19409  19109  19991  19709
27071  18547  27479  19403
53983  16583  54167  18433
96889  3553  96911  11353
108521  1447  108529  11503
160867  17699  160933  22501
170051  14173  170101  20123
171127  98783  171473  99817
212471  2833  212479  16217
261299  871  261301  17729
1085171  90433  1085221  97367
1724659  235709  1724741  240059
1906159  737093  1906391  738643
1926173  815467  1926427  816883
2156089  79057  2156111  94007
2991173  518479  2991277  521929
25257869  2609983  25257931  2615783
113476571  1513021  113476579  1557371
210972899  703243  210972901  864707
213602921  2848039  213602929  2892689
271181099  903937  271181101  1068887
995465099  3318217  995465101  3493583
7287489299  24291631  7287489301  24470969
24246030899  80820103  24246030901  80999903 

***

 

 

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