Ian H
Lv 7
Ian H asked in Science & MathematicsMathematics · 6 months ago

# Find the three integers a, b, c, such that are related by squares as:- a + b = x^2 b + c = y^2 a + c = z^2 a + b + c = w^2?

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• atsuo
Lv 6
6 months ago

*** 2nd edited ***

I found one method for solving this problem.

Problem : find positive integers a,b,c such that

a + b = x^2

b + c = y^2

a + c = z^2

a + b + c = w^2

a,b,c are positive, so x,y,z,w ≠ 0.

By given 4 equations, we can find

x^2 + y^2 + z^2 = 2w^2 ---(#1)

And if we can find x,y,z then a,b,c become

a = (1/2)(x^2 + z^2 - y^2) ---(#2)

b = (1/2)(x^2 + y^2 - z^2) ---(#3)

c = (1/2)(y^2 + z^2 - x^2) ---(#4)

To solve (#1), divide it by w^2.

(x/w)^2 + (y/w)^2 + (z/w)^2 = 2 ---(#5)

If (x/w)^2 ≧ 1 then (y/w)^2 + (z/w)^2 ≦ 1 so

(y/w)^2 + (z/w)^2 - (x/w)^2 ≦ 0

y^2 + z^2 - x^2 ≦ 0

c = (1/2)(y^2 + z^2 - x^2) ≦ 0 <--- unsuitable solution

So (x/w)^2 < 1. Similarly, (y/w)^2 < 1 and (z/w)^2 < 1.

x/w,y/w,z/w are rational numbers, so let them be X,Y,Z. It must be

0 < X^2,Y^2,Z^2 < 1 ---(#6)

Now, (#5) becomes

X^2 + Y^2 + Z^2 = 2 ---(#7)

*** parameters are simplified , 1/m = p and 1/n = q ***

To solve (#7), rewrite it with 2 parameters p and q. Let

X = pZ - 1 ---(#8)

Y = qZ - 1 ---(#9)

So (#7) becomes

(p^2Z^2 - 2pZ + 1) + (q^2Z^2 - 2qZ + 1) + Z^2 = 2

(p^2 + q^2 + 1)Z^2 - (2p + 2q)Z = 0 ---(#10)

We know Z ≠ 0, so (#10) becomes

(p^2 + q^2 + 1)Z - (2p + 2q) = 0

(p^2 + q^2 + 1)Z = 2(p + q)

Z = 2(p + q)/(p^2 + q^2 + 1)  ---(#11)

And we can find X and Y by (#8) and (#9).

X = 2p(p + q)/(p^2 + q^2 + 1) - 1 ---(#12)

Y = 2q(p + q)/(p^2 + q^2 + 1) - 1 ---(#13)

If p and q are rational numbers then X,Y,Z become rational numbers.

So we can choose any rational p and q as long as (#6) is satisfied.

*** add the conditions for parameters ***

If p(p + q) ≦ 0 then 2p(p + q)/(p^2 + q^2 + 1) ≦ 0, so (#12) becomes

X = 2p(p + q)/(p^2 + q^2 + 1) - 1 ≦ -1. Therefore X^2 ≧ 1.

This does not satisfy (#6), so p(p + q) must be positive.

Similarly, we can find that q(p + q) must be positive.

But if p and q are the combination of positive and negative numbers

then at least one of p(p + q) or q(p + q) is non-positive.

Therefore, p and q are both positive or both negative.

On the other hand, if we substitute -p and -q instead of p and q for

(#11)～(#13) then we get the same solution of (X^2,Y^2,Z^2).

Therefore, we can think that p and q are positive.

The condition for Z :

Z = 2(p + q)/(p^2 + q^2 + 1) < 1

2p + 2q < p^2 + q^2 + 1

1 < (p^2 - 2p + 1) + (q^2 - 2q + 1)

1 < (p - 1)^2 + (q - 1)^2 ---(#14)

The condition for X :

-1 < X = 2p(p + q)/(p^2 + q^2 + 1) - 1 < 1

0 < 2p(p + q)/(p^2 + q^2 + 1) < 2

2p^2 + 2pq < 2(p^2 + q^2 + 1)

p^2 + pq < p^2 + q^2 + 1

pq < q^2 + 1

p < q + 1/q ---(#15)

Similarly, the condition for Y :

q < p + 1/p ---(#16)

(#14)～(#16) are the conditions for p and q. They can be any positive

rational numbers.

Example : (p,q) = (2,1/3)

X = 2(2)(2 + 1/3)/(4 + 1/9 + 1) - 1 = 19/23

Y = 2(1/3)(2 + 1/3)/(4 + 1/9 + 1) - 1 = -16/23

Z = 2(2 + 1/3)/(4 + 1/9 + 1) = 21/23

x = 19

y = -16

z = 21

Check : 19^2 + (-16)^2 + 21^2 = 2 * 23^2

a = (1/2)(x^2 + z^2 - y^2) = 273

b = (1/2)(x^2 + y^2 - z^2) = 88

c = (1/2)(y^2 + z^2 - x^2) = 168

- The end -

• Ian H
Lv 7
6 months ago

a + b = x^2

b + c  = y^2

a + c = z^2

a + b + c = w^2

NOT A SOLUTION Just some early musings ....

2*t^2 is always a sum of three squares ?

x^2 + y^2 + z^2 = 2w^2,  ...suggets using known triples, as with

3^2 + 4^2 + 5^2 = 2*5^2

5^2 + 12^2 + 13^2 = 2*13^2

7^2 + 24^2 + 25^2 = 2*25^2

But that would use w = z forcing b = 0, but integers so zero not allowed

Here is a short list of numbers that are the sum of three nonzero squares.

3, 6, 9, 11, 12, 14, 17, 18, 19, 21, 22, 24, 26, 27, 29, 30, 33, 34, 35, 36, 38, 41, 42, 43, 44, 45, 46, 48, 49, 50, 51, 53, 54, 56, 57, 59, 61, 62, 65, 66, 67, 68, 69, 70, 72, 73, 74, 75, 76, 77, 78, 81, 82, 83, 84, 86, 88, 89, 90, 91, 93, 94, 96, 97, 98, 99, 101, 102, 104.

Some results that are also squares are

2^2 + 4^2 + 4^2 = 36,

2^2 + 3^2 + 6^2 = 49,

1^2 + 4^2 + 6^2 = 81

3^2 + 6^2 + 6^2 = 81

N. Beguelin: Every positive integer which is neither of the form 8n + 7, nor of the form 4n, is the sum of three squares.

Legendre:  A positive integer is sum of three squares unless it is of the form 4^a*(8b+7).

Anyone know how to approach this?

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