Find the minimum values of $a,b,c,d,e,f$ that satisfy following equations

Question

${ a }^{ 2 }+{ b }^{ 2 }={ c }^{ 2 }\\ { a }^{ 2 }+{ \left( b+c \right) }^{ 2 }={ d }^{ 2 }\\ { a }^{ 2 }+{ \left( b+c+d \right) }^{ 2 }={ e }^{ 2 }\\ { a }^{ 2 }+{ \left( b+c+d+e \right) }^{ 2 }={ f }^{ 2 }$

where $a,b,c,d,e,f$ positive integers. Moreover, prove that we can always find positive integer solutions when the number of equations goes to infinity, i.e.

${ a }^{ 2 }+{ b }^{ 2 }={ c }^{ 2 }\\ { a }^{ 2 }+{ \left( b+c \right) }^{ 2 }={ d }^{ 2 }\\ { a }^{ 2 }+{ \left( b+c+d \right) }^{ 2 }={ e }^{ 2 }\\ { a }^{ 2 }+{ \left( b+c+d+e \right) }^{ 2 }={ f }^{ 2 }\\ \vdots $

Note: You probably see this problem for the first time here.

Edit: A small example:

${ 24 }^{ 2 }+{ 7 }^{ 2 }={ 25 }^{ 2 }\\ { 24 }^{ 2 }+{ \left( 25+7 \right) }^{ 2 }={ 40 }^{ 2 }$

Answer 1

We consider the rational solutions first,and then multiply by an integer to make them be integers.

Assume $a=1,$ then $1+b^2=c^2,(c-b)(c+b)=1,$ let $c+b=t_1,c-b=\dfrac{1}{t_1},$ then we get $$c=\frac{1}{2}(t_1+\frac{1}{t_1}),b=\frac{1}{2}(t_1-\frac{1}{t_1}),$$ we need $1+(b+c)^2=1+t_1^2=d^2,$in the same way,we get $$d=\frac{1}{2}(t_2+\frac{1}{t_2}),t_1=\frac{1}{2}(t_2-\frac{1}{t_2}),$$ $$e=\frac{1}{2}(t_3+\frac{1}{t_3}),t_2=\frac{1}{2}(t_3-\frac{1}{t_3}),$$ $$f=\frac{1}{2}(t_4+\frac{1}{t_4}),t_3=\frac{1}{2}(t_4-\frac{1}{t_4}),$$ $$\cdots$$

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For example, let $t_4=11,$ then $t_3=\frac{60}{11},t_2=\frac{3479}{1320},t_1=\frac{10361041}{9184560},b=\frac{22995028210081}{190323205453920}.$

Then multiply by $190323205453920,$ we get $a=190323205453920,b=22995028210081,c=191707312997281,d=286914652560962,e=536509581434876,f=1055428684789920.$

You must choose $t_4$ so that $t_3,t_2,t_1,b>0$ .

Answer 2

I've used the Python interface of the Z3 solver to find solutions.

For the a, b, c case:

a, b, c = Ints('a b c')
s = Solver()
s.add(a > 0, b > 0, c > 0)
s.add(a * a + b * b == c * c)
#  additional constraint to look for "smaller" solutions
s.add(c < 6)
print s.check()
m = s.model()
print "a = %s" % m[a]
print "b = %s" % m[b]
print "c = %s" % m[c]

a = 3
b = 4
c = 5

For the a, b, c, d case the solver returns "unknown". The Microsoft Solver Foundation solver is also unable to find a solution for a, b, c, d (value range 1..5000).