(a,b,c); ab+1, ac+1, bc+1 and abc+1 are squares

 

Find positive integers   (a, \; b, \; c)   such that

a \, b \; + \; 1
a \, c \; + \; 1
b \, c \; + \; 1
a \, b \, c \; + \; 1

are all square numbers.

 

Here are two examples:

Example #1 :

a = 5,    b = 7,    c = 24

a \, b \; + \; 1 \; = \; (5\times 7) \; + \; 1 \; = \; 6^2
a \, c \; + \; 1 \; = \; (5\times 24) \; + \; 1 \; = \; 11^2
b \, c \; + \; 1 \; = \; (7\times 24) \; + \; 1 \; = \; 13^2
a \, b \, c \; + \; 1 \; = \; (5\times 7\times 24) \; + \; 1 \; = \; 29^2

we can express as:

(6^2 - 1) \,(11^2 - 1) \,(13^2 - 1) \; = \; (29^2 - 1)^2

 

Example #2 :

a = 45,    b = 8,    c = 91

a \, b \; + \; 1 \; = \; (45\times 8) \; + \; 1 \; = \; 19^2
a \, c \; + \; 1 \; = \; (45\times 91) \; + \; 1 \; = \; 64^2
b \, c \; + \; 1 \; = \; (8\times 91) \; + \; 1 \; = \; 27^2
a \, b \, c \; + \; 1 \; = \; (45\times 8\times 91) \; + \; 1 \; = \; 181^2

(19^2 - 1) \,(64^2 - 1) \,(27^2 - 1) \; = \; (181^2 - 1)^2

 

Find other solutions

 
 
Paul’s solutions:

{3,133,176},   {8,105,171},   {11,105,184},
{20,84,186},   {40,119,297},   {44,102,280}

 

(3\times 133) \; + \; 1 \; = \; 20^2
(3\times 176) \; + \; 1 \; = \; 23^2
(133\times 176) \; + \; 1 \; = \; 153^2
(3\times 133\times 176) \; + \; 1 \; = \; 265^2

(8\times 105) \; + \; 1 \; = \; 29^2
(8\times 171) \; + \; 1 \; = \; 37^2
(105\times 171) \; + \; 1 \; = \; 134^2
(8\times 105\times 171) \; + \; 1 \; = \; 379^2

(11\times 105) \; + \; 1 \; = \; 34^2
(11\times 184) \; + \; 1 \; = \; 45^2
(105\times 184) \; + \; 1 \; = \; 139^2
(11\times 105\times 184) \; + \; 1 \; = \; 461^2

(20\times 84) \; + \; 1 \; = \; 41^2
(20\times 186) \; + \; 1 \; = \; 61^2
(84\times 186) \; + \; 1 \; = \; 125^2
(20\times 84\times 186) \; + \; 1 \; = \; 559^2

(40\times 119) \; + \; 1 \; = \; 69^2
(40\times 297) \; + \; 1 \; = \; 109^2
(119\times 297) \; + \; 1 \; = \; 188^2
(40\times 119\times 297) \; + \; 1 \; = \; 1189^2

(44\times 102) \; + \; 1 \; = \; 67^2
(44\times 280) \; + \; 1 \; = \; 111^2
(102\times 280) \; + \; 1 \; = \; 169^2
(44\times 102\times 280) \; + \; 1 \; = \; 1121^2

 

pipo’s solutions:

{24,301,495},   {24,477,715},   {85,672,1235},
{114,816,1540},   {132,820,1610},   {165,664,1491},   {280,546,1608}

 

(24\times 301) \; + \; 1 \; = \; 85^2
(24\times 495) \; + \; 1 \; = \; 109^2
(301\times 495) \; + \; 1 \; = \; 386^2
(24\times 301\times 495) \; + \; 1 \; = \; 1891^2

(24\times 477) \; + \; 1 \; = \; 107^2
(24\times 715) \; + \; 1 \; = \; 131^2
(477\times 715) \; + \; 1 \; = \; 584^2
(24\times 477\times 715) \; + \; 1 \; = \; 2861^2

(85\times 672) \; + \; 1 \; = \; 239^2
(85\times 1235) \; + \; 1 \; = \; 324^2
(672\times 1235) \; + \; 1 \; = \; 911^2
(85\times 672\times 1235) \; + \; 1 \; = \; 8399^2

(114\times 816) \; + \; 1 \; = \; 305^2
(114\times 1540) \; + \; 1 \; = \; 419^2
(816\times 1540) \; + \; 1 \; = \; 1121^2
(114\times 816\times 1540) \; + \; 1 \; = \; 11969^2

(132\times 820) \; + \; 1 \; = \; 329^2
(132\times 1610) \; + \; 1 \; = \; 461^2
(820\times 1610) \; + \; 1 \; = \; 1149^2
(132\times 820\times 1610) \; + \; 1 \; = \; 13201^2

(165\times 664) \; + \; 1 \; = \; 331^2
(165\times 1491) \; + \; 1 \; = \; 496^2
(664\times 1491) \; + \; 1 \; = \; 995^2
(165\times 664\times 1491) \; + \; 1 \; = \; 12781^2

(280\times 546) \; + \; 1 \; = \; 391^2
(280\times 1608) \; + \; 1 \; = \; 671^2
(546\times 1608) \; + \; 1 \; = \; 937^2
(280\times 546\times 1608) \; + \; 1 \; = \; 15679^2

 

Can you solve it analytically?

 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

5 Responses to (a,b,c); ab+1, ac+1, bc+1 and abc+1 are squares

  1. paul says:

    Here are a few more, format {a, b, c}

    {3,133,176},
    {8,105,171},
    {11,105,184},
    {20,84,186},
    {40,119,297},
    {44,102,280}

    Paul.

  2. pipo says:

    A few more:
    {24, 301, 495}
    {24, 477, 715}
    {85, 672, 1235}
    {114, 816, 1540}
    {132, 820, 1610}
    {165, 664, 1491}
    {280, 546, 1608}

    pipo.

  3. pipo says:

    I tried, from the Diophantine triples we know:
    a= m
    b = n*(m*n +2)
    c = (n+1)*(m*n +m +2)
    We get:
    ab+1 = (m*n+1)^2
    ac+1 = (m*n+ m +1)^2
    bc +1 = (m*n^2 + m*n + 2n + 1)^2

    But I am stuck with abc + 1 is a square.

    pipo

    • benvitalis says:

      Let me re-write your expressions as:
      a = n
      b = k*(k*n + 2)
      c = (k+1)*((k+1)*n + 2)

      Substitute the parameterization expressions for a,b,c
      into the equation a*b*c + 1 = x^2
      then you have a quartic equation in k

      we need k to be rational

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