Find positive integers such that

are all square numbers.

Here are two examples:

Example #1 :

a = 5, b = 7, c = 24

we can express as:

Example #2 :

a = 45, b = 8, c = 91

Find other solutions

Paul’s solutions:

{3,133,176}, {8,105,171}, {11,105,184},

{20,84,186}, {40,119,297}, {44,102,280}

pipo’s solutions:

{24,301,495}, {24,477,715}, {85,672,1235},

{114,816,1540}, {132,820,1610}, {165,664,1491}, {280,546,1608}

**Can you solve it analytically?**

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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.

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.

Good work guys! Can you solve it analytically?

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

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