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I am looking for the number of integer solutions to this system of equations

$$a^2+b^2+c^2+d^2=2500$$ and

$$(a+50)(b+50)=cd$$

I tried moving terms around in the first equation and using the difference of two squares to try and gain some information, but came up empty. I was wondering what the best way to attack this problem is. Should I take the equations mod something, or should I try and place some bounds on the variables and try to derive the number of solutions from that? I ask that y'all only give me hints. Thanks

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  • $\begingroup$ Hence you are entitled your question as "Competition Diophantine Equation" I want you to clarify something: Are we talking about and ongoing competition or is this task from a former year? $\endgroup$ – mrtaurho Sep 28 '18 at 18:05
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    $\begingroup$ This problem is from a previous AMATYC, a math competition for two year colleges. $\endgroup$ – L. Tim Sep 28 '18 at 19:15
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Hint: Use the fact $$2cd\leq c^2+d^2$$

and replace $cd$ and $c^2+d^2$ with a given constrains.

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Note that $$ \eqalign{(c-d)^2 &= c^2 + d^2 - 2 c d \cr &= 2500 - a^2 - b^2 - 2 (a+50)(b+50)\cr &= -(a+b+50)^2\cr}$$ Since a square must be nonnegative, we must have $c=d$ and $a+b=-50$. The second equation then becomes $a^2 + 50 a + c^2 = 0$. After completing the square, we see that solutions are related to the Gaussian integer factorization of $625$.

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  • $\begingroup$ This is a hint? $\endgroup$ – Aqua Sep 28 '18 at 20:15
  • $\begingroup$ Sorry, got carried away. $\endgroup$ – Robert Israel Sep 28 '18 at 20:17
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It is well known that by setting $$ r_4(n) = \left|\{(a,b,c,d)\in\mathbb{Z}^4: a^2+b^2+c^2+d^2=n\}\right| $$ we have $$ r_4(n) = 8\sum_{\substack{d\mid n\\ 4\nmid d}}d $$ for instance through Jacobi's triple product. In particular the number of ways for writing $n$ as a sum of four squares is a constant multiple of a multiplicative function, which only depends on the factorization of $n$. Since $2500 = 2^2\cdot 5^4$ we have $$\begin{eqnarray*} r_4(2500) &=& 8\left(1+5+5^2+5^3+5^4+2+2\cdot 5+2\cdot 5^2+2\cdot 5^3+2\cdot 5^4\right)\\&=&8\cdot 3\cdot\frac{5^5-1}{5-1}=6\cdot(5^5-1)=\color{red}{18744}.\end{eqnarray*} $$ In the given problem we have an extra constraint, and it is a heavy one, since it leads to $$ 2500 = a^2+b^2+c^2+d^2 = cd-ab-50(a+b),$$ $$ (a^2+ab+b^2)+50(a+b)+(c^2-cd+d^2) = 0, $$ and by setting $a=\frac{A-50}{3},b=\frac{B-50}{3}$ we get $$ (A^2+AB+B^2)+9(c^2-cd+d^2) = 7500,\quad A\equiv B\equiv 2\pmod{3}$$ which clearly has a finite number of solutions, since both $x^2\pm xy+y^2$ and $x^2+9y^2$ are positive definite quadratic forms. The integers represented by $x^2\pm xy+y^2$ are the ones given by a product of primes of the form $3k+1$ and squares of primes of the form $3k+2$. For a brute-force approach, one may simply list all the solutions of the above equation, then check which ones meet $a^2+b^2+c^2+d^2=2500$.

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  • $\begingroup$ Might writing $a^2\pm ab + b^2 = \frac12(a^2\pm 2ab+b^2+a^2+b^2)=\frac12((a\pm b)^2+a^2+b^2)$ help? $\endgroup$ – marty cohen Sep 28 '18 at 20:49
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Above system of equations is shown below:

$a^2+b^2+c^2+d^2=2500$

$(a+50)(b+50)=cd$

The numerical integer solution to the above is given below:

$(a,b,c,d)=(-25,-25,-25,-25)$

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I solve Robert Israel's equation. This lead pytagorean equation.

$(a+50)^2+c^2=25^2$

famous examples $3^2+4^2=5^2$,$7^2+24^2=25^2$,and trivial one $0^2+25^2=25^2$ are solutions. Final answers is $(a,b,c,d)=(-5,-45,±15,±15)(-10,-40,±15,±15)(-18,-32,±24,±24)(-1,-49,±7,±7)(0,-50,0,0)(-50,0,0,0)$

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