single quadratic diphontine equation with 4 unknowns, all solutions?

Question

I am trying to find a general solution for $d\left(a+b-c\right) = a^2+b^2-c^2$ with the requirement that all variables are integers.

Obviously $\frac{a^2+b^2-c^2}{a+b-c}$ must an integer, but my understanding is that there must be some constraints on the values of a, b, or c for this to assuredly be the case.

I easily discovered the trivial solution where $a = 1-b+c$, but I'm fairly sure a more general solution exists, albeit with constraints on more than just one unknown. How can I find it?

Answer 1

►Putting $2d$ instead of $d$ we have $(a-d)^2+(b-d)^2=c^2-2cd+2d^2$ so for $c$ and $d$ fixed, $a$ and $b$ are coordinates of points of the circle $(x-d)^2+(y-d)^2=c^2-2cd+2d^2$.

Example.-For $(c,d)=(1,4)$ we have the four integer points $(a,b)=(1,0),(3,0),(4,1),(4,3)$ and the symmetrics respect to the diagonal $y=x$

►►In general we have $(a-d)^2+(b-d)^2=(c-d)^2+d^2$ so each solution $(a,b,c,2d)$ correspond to a solution of the diophantine equation $x^2+y^2=z^2+w^2$ whose general solution is given by the known identity $$(mX+nY)^2+(-nX+mY)^2=(nX+mY)^2+(mX-nY)^2$$ where $m,n,X,Y$ are arbitrary integers.

Consequently we can put as parameterization of the solutions $$\begin{cases}a-d=mX+nY\\b-d=-nX+mY\\c-d=nX+mY\\d=mX-nY\end{cases}\Rightarrow \begin{cases}a=2mX\\b=(m-n)(X+Y)\\c=(m+n)X+(m-n)Y\\d=mX-nY\end{cases}$$ Example.-$(m,n,X,Y)=(3,1,5,11)$ gives $(a,b,c,d)=(30,32,42,4)$ but, keeping in mind that we have by confort put $2d$ instead of $d$ we have the equality of verification $$\frac{30^2+32^2-42^2}{30+32-42}=8$$ We find as quotient $8$ instead of $4$ but this is logic; this is so for any other example of $(m,n,X,Y)$ we chose because we have put $2d$ instead of $d$ at the beginning.

Answer 2

Given equation is:

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

Above has parametric solution:

$a=6(9-8k)$

$b=11(2k-1)$

$c=5(11-10k)$

$d=(12k-1)$

for, $k=2$ we have solution in integers:

$(a,b,c,d)=(-42,33,-45,23)$

Answer 3

Added: this is based on page 254, volume II of History of the Theory of Numbers by Dickson. Here Welsch gave the general solution of $$ U^2 + X^2 = Y^2 + Z^2 \; , \; \; $$ I applied that to your $$ (2a-d)^2 + (2b-d)^2 = (2c-d)^2 + d^2 \; \; . \; \; $$

He used parameters abcd so I had to fiddle with the letters. In turn, that can be proved by this: if $j,k,l,m,$ are nonzero integers and $jk=lm$ we may take $g = \gcd(j,l)$ to reach $$ j = g \alpha, k = h \beta, l = g \beta, m = h \alpha.$$ Here $\gcd(\alpha, \beta) = 1.$ If, in addition, $\gcd(j,k,l,m)=1$ then also $\gcd(g,h)=1$

Four integers $p,q,r,s$ with both $p,s$ even, or $q,r$ even.

This version came out better, simple $\pmod 2$ condition, either $p+r$ or $q+s$ must be even. $$a = pq $$ $$b = \frac{ pq - ps + qr - r s}{2} $$ $$ c = \frac{pq + ps + qr - r s}{2} $$ $$d = pq - rs $$

? 
?   a = p*q;  
b = p*q - p*s + q*r - r * s ;  b /= 2;  
c =   p*q + p*s + q*r - r * s;  c /= 2;  
d =  p*q - r*s; 
 num = a^2 + b^2 - c^2 ; 
 den = a+b - c;
? 
? 
? a
%2 = q*p
? b
%3 = (1/2*q - 1/2*s)*p + (1/2*r*q - 1/2*r*s)
? c
%4 = (1/2*q + 1/2*s)*p + (1/2*r*q - 1/2*r*s)
? d
%5 = q*p - r*s
? num
%6 = (q^2 - s*q)*p^2 + (-r*s*q + r*s^2)*p
? den
%7 = (q - s)*p
? d * den
%8 = (q^2 - s*q)*p^2 + (-r*s*q + r*s^2)*p
? d * den  - num
%9 = 0

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

I finally ran a big check with hypothesis $a+b - c \neq 0,$ furthermore reported quadruples $(a,b,c,d)$ with $\gcd(a,b,c,d)=1.$ I printed when $a^2 + b^2 + c^2 + d^2 \leq \color{red}{1500}. \; \;$ Quite a number of these. I had it run the $pqrs$ business with all four bounded in absolute value; when this bound was set to $\color{red}{38}$ all the $abcd$ quadruples were produced. When I decreased this bound to $37$
a few quadruples were missed, either $a=c$ and $ b = \pm d$ or $b=c$ and $ a = \pm d$

  STILL MISSING    
       -21        17       -21        17
       -21        17        17       -21
       -17        21       -17        21
       -17        21        21       -17
        17       -21       -21        17
        17       -21        17       -21
        21       -17       -17        21
        21       -17        21       -17
  bound0 was   1500
  bound was   37


------------------------------


  STILL MISSING    
  bound0 was   1500
  bound was   38

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

Answer 4

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

Eqn (1) has parametric solution shown below:

$a=(m+n)(m-n)$

$b=(2mn)$

$c=n(m-n)$

$d=(m^2-mn+2n^2)$

For, $(m,n)=(3,2)$ we have:

$(a,b,c,d)=(5,12,2,11)$

"OP" in his comments said that he is looking for a genral solution to

eqn (1). FYI, general solutions are quite rare for Diophantine

equations. Also it is quite difficult to prove that a parametric

solution found is a general solution. So all the best in his

quest in looking for a general solution.