$a+b=u^2, a^2+b^2=v^4$
I have found the solution: $a=4565486027761, \quad b=1061652293520, \quad u=2372159, \quad v=2165017$.
But I do not know a more theoretical way to get them.
$a+b=u^2, a^2+b^2=v^4$
I have found the solution: $a=4565486027761, \quad b=1061652293520, \quad u=2372159, \quad v=2165017$.
But I do not know a more theoretical way to get them.
Above simultaneous equations shown below;
$a+b=u^2$, $a^2+b^2=v^4$
I think "OP" has rediscovered numerical solutions to above equation first given by Fermat. The solution is given on page 621 in the book by "L E Dickson", Vol. (2) "History of theory of numbers".
Euler took $a=(r^4-6r^2s^2+s^4)$, $b=(4r^3s-4rs^3)$ & $v=(r^2+s^2)$
We get:$u=(1/2)(115s^2-80rs+2r^2)$
For $(r,s)=(1469,84)$ the above numerical solution is obtained
Regarding above equation's shown below:
$a+b=u^2$ and
$a^2+b^2=v^4$
Solution is shown below:
$(a,b,u,v)=[(120), (-119), (1), (13)]$
There is another solution given on Seiji Tomita site and the link is shown below.
http://www.maroon.dti.ne.jp/fermat
And click on computational number theory and then click on article # 146
The numerical solution is shown below:
$a=214038981475081188634947041892245670988588201$
$b=109945628264924023237017010068507003594693720$
$u= 17999572487701067948161$
$v=15512114571284835412957$
This is a classic problem of Fermat as mentioned in another answer. Use a Google search on $4565486027761$ to confirm it. I have found a method to generate all solutions to the Diophantine problem which uses generalized Somos-4 sequences. Define the following six sequences:
$$ x_0 = 0,\, x_1 = 1,\, x_2 = -3,\, x_n = x_{-n},\, y_n = -y_{-n}, \\ y_0 = 0,\, y_1 = 1,\, y_2 = 2,\, y_3 = -13,\, y_4 = 84,\, y_5 = 1525, \\ x_n = (-13x_{n-1}x_{n-4} - 42x_{n-2}x_{n-3})/x_{n-5},\\ y_n = (-13y_{n-1}y_{n-4} - 42y_{n-2}y_{n-3})/y_{n-5},\\ z_n := x_{n-1}y_{n+1} - x_{n+1}y_{n-1}, \\ a_n := (x_n^2 + z_n)/2, \;\; b_n := (x_n^2 - z_n)/2, \;\; c_n := \sqrt{a_n^2 + b_n^2}. $$
A small table of values: $$\begin{array}{|c|c|c|c|c|c|c|} \hline n & a & b & c & x & y & z \\ \hline 1 & 1 & 0 & 1 & 1 & 1 & 1 \\ \hline 3 & -119 & 120 & 169 & 1 & -13 & -239 \\ \hline 5 & 2276953 & -473304 & 2325625 & -1343 & 1525 & 2750257 \\ \hline \end{array}$$
All of them are integer sequence except $c_n$ for even $n$. In the question example:
I have found the solution: $a=4565486027761, \quad b=1061652293520, \quad u=2372159, \quad v=2165017$.
the solution given is identified as: $$ a = a_7,\;\; b = b_7,\;\; u = -x_7,\;\; v = -y_7. $$
One theoretical way to obtain this solution is using Pythagorean triples $(a,b,c)$ with the additional requirements that $a+b$ is a perfect square and $a^2+b^2$ is a fourth power; there is a formula for generating (primitive) Pythagorean triples. The triple for the above solution is $$ 4565486027761^2+1061652293520^2=4687298610289^2 $$