Can you find four numbers such that the sum of every two numbers and the sum of all four numbers may be perfect squares???
Answers
Step-by-step explanation:
0,0,0,A works for A any perfect square. More generally, 0,0 together with any of the small pair of squares in a pythagorean triple, e.g. (0,0,9,16). More generally still, for any pair of integers m and n,
(2m^2 n^2 , 2m^2 n^2 , 2m^2 n^2 , (m^2 - n^2)^2 - 2 m^2 n^2)
is a solution. For example, with m=2, n=1,
(8,8,8,1)
More solutions are
2m^2 n^2, 2m^2 n^2 , - 2m^2 n^2 , (m^2 - n^2)^2 - 2 m^2 n^2
for example,
8,8,-8,17
Now only two of the numbers are the same. So I assume you mean all four numbers are positive and different.
The basic theorem to know to solve this in general is that all pythagorean triples are of the form 2mn,m^2-n^2, (m^2+n^2)^2, which is proved simply from the defining equation and some unique factorization/Euclid's algorithm considerations (this is a classical result).
Given this, the sum of all the numbers is a pythagorean triple by each of the three possible pairings of the four numbers. If there is a unique decomposition for the long-leg of the triangle into integer short legs, the answer is one of the trivial cases above.
But not all squares are uniquely decomposable into smaller squares. The method for finding the number of compositions is through the Gaussian integers: ab has the same length in the Gaussian integers as ab*, and from unique factorization in the Gaussian integers, this is the general method to generate numbers with the same length. For example, using a = 2+3i and b = 3+i, you get the square identity:
33^2 + 56^2 = 16^2 + 63^2
This then generates the following family:
x, 33^2 - x , 16^2 - x , 56^2 - 16^2 + x
which, by the parametrization, autmatically makes the sum of all 4 a square, and the decomposition of 1+2, 1+3, 2+3 squares.
The nontrivial condition will work the moment 56^2 - 16^2 + 2x is one of the options 56^2, 33^2, 16^2 or 63^2. I am not sure if any of these give nontrivial positive solutions, I didn't bother checking, because by finding larger and larger sets of Gaussian integers all sharing the same length, you can generate as many nontrivial solutions as you like.
For example, suppose you find three Gaussian integers with the same length:
m^2 + n^2 = p^2 + q^2 = s^2 + t^2
then
(x, (2mn)^2 -x, (2pq)^2 -x , (m^2 - n^2) - (2pq)^2 +x)
will work, so long as you choose x so that
(m^2 - n^2)^2 - (2pq)^2 + 2x = (s^2 - t^2)^2
or (2st)^2 on the right hand side. This method generates all the solutions with the constraints provided.