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I have reduced a certain equation (in positive integers) to the equation

$$A^2 + B^2 + 4 = C^2 + D^2. \quad(\star)$$

Assume the positive integers $(a,b,c,d)$ are any solution to $(\star)$. Are there any algebraic restrictions on [i.e., relationships between] the elements which are well-known or easy to prove? I'm thinking of things like Pythagorean means, or other relative size or congruence restrictions.

Thank you,
Kieren.

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One special case $D=2\implies A,B,C $ are Pythagorean triple (en.wikipedia.org/wiki/Pythagorean_triple) –  lab bhattacharjee Sep 15 '13 at 17:03

5 Answers 5

One can pick up some things from congruence considerations. Here is a sample. Let us work modulo $8$.

If $a$ and $b$ are odd, the left side is congruent to $6$ modulo $8$. But the right side cannot be congruent to $6$ modulo $8$. (We used the fact that if $x$ is odd, then $x^2\equiv 1\pmod{8}$.)

Suppose now that $a$ is even and $b$ is odd. If $a$ is divisible by $4$, then $a^2+b^2+4$ is congruent to $5$ modulo $8$, which means that one of $c$ or $d$ is odd, and the other is even but not divisibl by $4$.

Let $a$ and $b$ be both even. Then $c$ and $d$ also must be. We end up with an equation of shape $s^4+t^4 +1=u^2+v^2$.

We could also pick up information by working modulo $3$, and other moduli.

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Thanks! One point: the right-hand side can be congruent to $5$ modulo $8$, if e.g. $2 \parallel c$ and $d$ is odd. –  Kieren MacMillan Sep 15 '13 at 18:29
    
Thanks, that was careless. –  André Nicolas Sep 15 '13 at 18:37
    
By the way, this was already quite helpful: using my reformulation and your congruence suggestion, I've already been able to eliminate one of the two possible cases I need to prove! Hopefully your "working modulo 3 and other moduli" will prove equally helpful. –  Kieren MacMillan Sep 15 '13 at 23:52
    
I am not optimistic about general solution. However, one should be able to get a parametric family of solutions. –  André Nicolas Sep 16 '13 at 0:01

A subset of integer solutions to,

$$A^2+B^2+4 = C^2+D^2\tag{1}$$

can be reduced to solving the Pell equation $x^2-ny^2 = 1$ for any non-square $n$ (where $n=5$ will involve the Fibonacci numbers). Consider the more general,

$$x_1^2+x_2^2+x_3^2 = y_1^2+y_2^2+y_3^2\tag{2}$$

The complete solution is,

$$(a+b)^2+(c+d)^2+(e+f)^2 = (a-b)^2+(c-d)^2+(e-f)^2\tag{3}$$

where,

$$ab+cd+ef = 0\tag{4}$$

It is easy to make the last term of $(3)$ vanish. Let $e=f=t$. Thus,

$$(a+b)^2+(c+d)^2+(2t)^2 = (a-b)^2+(c-d)^2\tag{5}$$

However, if you set $a,b,c = x+y,\, -x+y,\, y^2$, then condition $(4)$ becomes,

$$x^2-(d+1)y^2 = t^2\tag{6}$$

If you set $t = 1$, then any Pell equation will give integer solutions to $(1)$. For a nice example, let $d = 4$ so one has to solve,

$$x^2-5y^2 = 1$$

The solution is $x_n = \tfrac{1}{2}F_{6n}+F_{6n-1}$ (A023039) and $y_n = \tfrac{1}{2}F_{6n}$ (A060645), where $F_n$ are the Fibonacci numbers. After some tweaking so that all terms won't be even, a Fibonacci identity which satisfies $(1)$ is then,

$$(F_{6n+3})^2+( \tfrac{1}{4}F_{6n+3}^2+4)^2+4 = (F_{6n+2}+F_{6n+4})^2+(\tfrac{1}{4}F_{6n+3}^2-4)^2\tag{7}$$

where $n=1$ is,

$$34^2+293^2+4 = 76^2 +285^2$$

and so on.

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I'm not sure you can say a great deal. If $A^2+B^2=C^2$ (pythagoren triple), then we can have $D=2$.

We have $4=(C+A)(C-A)+(D+B)(D-B)$ so that $C+A$ and $D+B$ are both odd or both even.

In the even case, both products are divisible by $4$ and we have some expression like $4=56-52=14\times 4 - 26\times 2$ which gives $C=9, A=5, B=14, D=12$

In the odd case we get something like $4=55-51=5\times 11-3\times 17$ which gives $C=8, A=3, D=7, B=10$

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The conditions for a number $n$ to be a sum of two squares are well-understood: $n$ must have no factors congruent to $3\bmod 4$ which occur to odd powers; in other words, if $p\equiv 3\pmod 4, p^{2k+1}|n$, then $p^{2k+2}|n$ (and in particular, if $p|n$ then $p^2|n$). This is a direct consequence of Fermat's Two-Square Theorem and the rule $(a^2+b^2)(c^2+d^2) = (ac-bd)^2+(ad+bc)^2$ for expressing the product of two numbers, each a sum of two squares, as a sum of two squares.

Applied to the given problem, this says that both $n=A^2+B^2$ and $n+4$ have no factors of their squarefree parts congruent to $3\bmod 4$; unfortunately, since addition plays notoriously poorly with factorization (for instance, beyond the trivial lack of shared factors, there's essentially nothing known about any correlation between the factorizations of $n$ and $n+1$) then it's unlikely that you'll be able to say anything about $A,B,C,D$ aside from the simplest congruence implications.

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For the equation: $X^2+Y^2+4=Z^2+D^2$

You can draw a simple formula.

$X=(2c+b+2)c+b$

$Y=(2c+b+2)c$

$Z=2c^2+bc-2$

$D=(2c+b+4)c+b$

more:

$X=\frac{(a^2-q^2)b}{2}+(b-2)(q-1)$

$Y=ab$

$Z=bq+2-b$

$D=\frac{(a^2-q^2)b}{2}+(b-2)q+2$

more:

$X=\frac{(a^2-q^2)b}{2}-(b+2)(q+1)$

$Y=ab$

$Z=b+bq+2$

$D=\frac{(a^2-q^2)b}{2}-(b+2)q-2$

$c,b,q,a$ - what some integers.

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