Say $(a+b)^(2)+1$ $=$ $(c+d)^2$ So what are the values of c and d?

Question

If we say $z=a^2+b^2$

then we get
$z+2ab+1=c^2+d^2+2cd$

and Say $w=c^2+d^2$

The equation is $z+2ab+1=w+2cd$

Subtract $w$ and $[2ab+1]$

$z-w=2cd-2ab-1$

since $2cd=(c+d)^2-c^2-d^2$

Witch is $2cd=(c+d)^2-w$

that simplifies down to

$2cd=(a+b)^2+1-w$

Do the same for $2ab$ and get

$2ab=(c+d)^2-1-z$

This is as far i can go.

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Answer 1 · 2018-02-18T08:57:33

The only solutions in non-negative integers are:

$(a, b, c, d) = (0, 0, 1, 0)$

and:

$(a, b, c, d) = (0, 0, 0, 1)$

Unless there are additional constraints on $a, b, c, d$ beyond what we have been told in the question then about all we can say is:

$c+d = +-sqrt(a^2+2ab+b^2+1)$

So you could solve for $c$ as:

$c = -d+-sqrt(a^2+2ab+b^2+1)$

or for $d$ as:

$d = -c+-sqrt(a^2+2ab+b^2+1)$

If $a, b, c, d$ are all integers then we are looking for two integer squares that differ by $1$ . The only pair is $1, 0$ .

Hence we find:

$(a+b)^2 = 0$

$(c+d)^2 = 1$

So:

$c+d = +-1$

So we could write:

$c = -d+-1$

$d = -c+-1$

Alternatively, if $a, b, c, d$ are all non-negative integers then this reduces the possible set of solutions to:

$(a, b, c, d) in { (0, 0, 1, 0), (0, 0, 0, 1) }$