Sum of Four Squares

$a^2 + b^2 + c^2 + d^2 = 2b^2 + c^2 - 2cd$ can be rearranged to:

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

and $a^2 + b^2 + c^2 + d^2 = b^2 + 2c^2 - 2ab$ can be rearranged to:

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

Now consider the following diagram of a rectangle with sides $a + b$ and $c + d$ :

According to the equations above, by Pythagorean's Theorem each of the three right triangles' hypotenuses are equal to each other, so the center triangle must be an equilateral triangle. Let the side of that equilateral triangle be $s$ .

By Pythagorean's Theorem $b^2 + c^2 = s^2$ , so if $a$ , $b$ , $c$ , and $d$ are positive integers, then $s^2$ must be an integer, too, and by the area of an equilateral triangle $A = \frac{\sqrt{3}}{4}s^2$ the area $A$ must be irrational.

However, the area of the equilateral triangle is the area of the rectangle minus the sum of the three right triangles, or $A = (a + b)(c + d) - (\frac{1}{2}(a + b)d + \frac{1}{2}bc + \frac{1}{2}a(c + d)) = \frac{1}{2}(ac + bc + bd)$ , and if $a$ , $b$ , $c$ , and $d$ are positive integers, then the area $A$ must be rational.

The area can't be rational and irrational at the same time, so it can't be that $a$ , $b$ , $c$ , and $d$ are all positive integers.

Therefore, there are zero solutions to the given equations with the given conditions.

The equations can be rewritten as $a^2 + (c+d)^2 \; = \; b^2 + c^2 \; =\; (a+b)^2 + d^2$ If we suppose that the HCF of $b$ and $c$ is $u$ , so that $b = uB$ , $c = uC$ for coprime positive integers $B,C$ , then $uC(c+2d) \; = \; (c+d)^2 - d^2 \; =\; (a+b)^2 - a^2 \; = \; (2a+b)uB$ so that $c + 2d = vB$ , $2a+b = vC$ for some positive integer $v$ . But then $v^2(B^2+C^2) \; = \; b^2 + c^2 \; = \; a^2 + (c+d)^2 \; =\; \left(\tfrac12(vC - uB)\right)^2 + \left(\tfrac12(vB + uC)\right)^2 \; = \; \tfrac14(u^2+v^2)(B^2+C^2)$ and hence $v^2 = \tfrac14(u^2+v^2)$ , and hence $u^2 = 3v^2$ . Thus, if there were any positive integer solutions of this equation, then $\sqrt{3}$ would be rational.

Thus there are no solutions.

Take twice the first equation and subtract out the second and third from it. This gives $2a^2+2b^2+2c^2+2d^2-(2b^2+c^2-2cd)-(b^2+2c^2-2ab)=0$ $2a^2+2d^2-b^2-c^2+2ab+2cd=0$ $2a^2+2d^2=b^2+c^2-2ab-2cd$ $3a^2+3d^2=a^2+b^2+c^2+d^2-2ab-2cd=(a-b)^2+(c-d)^2$ Now squares are only 0 or 1 mod 3 and since the LHS is divisible by 3, $a-b\equiv 0 \pmod{3}\quad c-d\equiv 0 \pmod{3}$ However due to the nature of squares having even prime exponents, this would actually make each of the squares on the RHS be divisible by 9! This would force $a^2\equiv 0 \pmod{3} \quad d^2\equiv 0 \pmod{3}$ $a\equiv 0 \pmod{3} \quad d\equiv 0 \pmod{3}$ $b\equiv a\equiv 0\pmod{3} \quad c\equiv d\equiv 0 \pmod{3}$ so all of the variables are divisible by 3, violating the gcd restriction. Thus there are no solutions

a²+b²+c²+d²=2b²+c²-2cd => b²-a²=2cd+d²...(i) Comparing 1st and 3rd part of given eqn, we get, b²+2ab=d²-c²...(ii) By (i)-(ii) we get, -a²-2ab=c²+2cd, which is impossible if a,b,c,d are positive integers

1. b2 > 2cd as a2+d2 > 0
2. c2 > 2ab as a2+d2 > 0 So b2 - c2 = 2(cd-ab) can't be true