Angeline Hayden

2022-04-22

Parametric solutions for the quadratic Diophantine equation $2{a}^{2}+{b}^{2}=2{c}^{2}+{d}^{2}$ .

silviavalls1005r

Beginner2022-04-23Added 11 answers

Write the equation as

$2({a}^{2}-{c}^{2})={d}^{2}-{b}^{2}$

i.e.$2(a+c)(a-c)=(d+b)(d-b)$

Now$d+b\text{}\text{and}\text{}d-b$ must be both even (they must have the same parity since $d=\frac{(d+b)+(d-b)}{2}$ is an integer, and they can't be both odd because $2(a+c)(a-c)$ is even. Then $(a+c)(a-c)$ is even, so $a+c\text{}\text{and}\text{}a-c$ are both even. We can write $a+c=2xy,a-c=2zw,d+b=4xz,d-b=2yw$ , so $a=xy+zw,c=xy-zw,d=2xz+yw,b=2xz-yw$ , or else $d+b=2yw\text{}\text{and}\text{}d-b=4xz$ , which produces a solution that is the same except for $b=-2xz+yw$ .

i.e.

Now

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