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$a+b=u^2, a^2+b^2=v^4$

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.

Qiuye Jia
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    How did you find your solutions? Please show us your work, else we cannot point you to what might be a "more theoretical" way to find them. As it is, it seems you reported solutions from a solutions manual and claimed to have found the solutions yourself. And have asked us to do the work for you. Again, if that's not the case, show us your work the led you to find the solutions you claim you found. – amWhy Nov 16 '18 at 01:17
  • I know it sounds funny... But I really forgot how I got them... I guess by computer... some semi-brutal way... – Qiuye Jia Jul 12 '23 at 06:08

4 Answers4

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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

Sam
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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:

$a=214038981475081188634947041892245670988588201$

$b=109945628264924023237017010068507003594693720$

$u= 17999572487701067948161$

$v=15512114571284835412957$

Sam
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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: $$ a = a_7,\;\; b = b_7,\;\; u = -x_7,\;\; v = -y_7. $$

Somos
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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 $$ 4565486027761^2+1061652293520^2=4687298610289^2 $$

Dietrich Burde
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