## (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?

math grad - Interest: Number theory
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### 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.

• benvitalis says:

Good work guys! Can you solve it analytically?

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