Daniel's equation

It is well-known that $x^2\equiv 0,1\pmod 4$ Let us consider $d>1$ . Therefore, $a^2\equiv b^2\equiv c^2\equiv 0 \pmod 4$ and $a\equiv b\equiv c\equiv 0 \pmod 2$ If $d>1$ , we can divide both sides by 4. Let $a=2x$ $b=2y$ $c=2z$ Now, we have $x^2+y^2+z^2=2^{d-2}$ Note that this is identical to our original equation. We can continue reducing it until the exponent on the RHS becomes 0 or 1. Let the new equation be $p^2+q^2+r^2=2^s$ We have two cases.

Case 1: $s=0$ and therefore $d\equiv 0\pmod 2$

In this case, $p^2+q^2+r^2=1$ Therefore, one of $(p,q,r)$ is 1, and the other two are 0. This gives 3 solutions, and since each even $d$ comes down to this case, there are $3\cdot 50=150$ solutions in this case.

Case 1: $s=1$ and therefore $d\equiv 1\pmod 2$

In this case, $p^2+q^2+r^2=2$ Therefore, one of $(p,q,r)$ is 0, and the other two are 1. This gives 3 solutions, and since each odd $d$ comes down to this case, there are $3\cdot 50=150$ solutions in this case.

Therefore, there are 300 solutions.

Consider the case $d = 0$ , we have: $a^2 + b^2 + c^2 = 1$ which has solutions, the $3$ permutations of $(0,0,1)$ . For $d=1$ , we similarly have $(a,b,c)$ a permutation of $(1,1,0)$ , which are $3$ in number. I claim all rest values of $d$ can be mapped on these two cases using descent .

For $d \geq 2$ , $a^2 + b^2 + c^2 = 2^d$ . Considering $\pmod{4}$ , we have all a,b & c are even. Let $a=2a_1,b=2b_1,c=2c_1$ , then we get $a_1^2 + b_1^2 + c_1^2 = 2^{d-2}$ .

If $d-2=0$ ,then $(a_1,b_1,c_1) \equiv (0,0,1)$ in some order $\Rightarrow (a,b,c) \equiv (0,0,2)$ in some order. If $d-2=1$ ,then $(a_1,b_1,c_1) \equiv (0,1,1)$ in some order $\Rightarrow (a,b,c) \equiv (0,2,2)$ in some order.

If $(d-2) \geq 2$ , then we might get as above, $(d-4) = 0$ or $1$ ,then get $(a_2,b_2,c_2) \equiv (0,0,1)$ or $(0,0,1)$ ,thus giving 3 solutions in each case.

It is easily noticed that after each division by 4, the equation for $d$ becomes same as the equation for $(d-2)$ . Continuing in this way, for each $d$ from $0$ onward, we get three solutions. Since $d < 100$ , we have 300 solutions, 3 for each $0 \leq d \leq 99$ .

The problem hinges on one interesting fact, which we prove as a lemma.

Lemma If any one of $a,b,c$ is odd, then $d<2$ .

Proof First, we note that if any one of $a,b,c$ is odd, then exactly two are unless $d=0$ . Otherwise, the parity of the LHS is not the same as the parity of the RHS.

WLOG, let $a=2x,b=2y+1,c=2z+1$ for some nonnegative integers $x,y,z$ . We expand $a^2+b^2+c^2$ to get $4x^2+4y^2+4y+1+4z^2+4z+1 \equiv 2 \pmod{4}$ . But note that if $d \geq 2, 2^d \equiv 0 \pmod{4}$ . Therefore, $d<2$ . QED

We quickly check and find that for $d=0, (a,b,c)$ is a permutation of $(1,0,0)$ , giving three possibilities, and for $d=1, (a,b,c)$ is a permutation of $(1,1,0)$ , again giving three possibitilities.

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We now note that if $(a,b,c)=(2x,2y,2z)$ for some nonnegative integers $x,y,z$ and $a^2+b^2+c^2=2^d$ , then $x^2+y^2+z^2 = 2^{d-2}$ . Now, either $d-2<2$ and $(x,y,z)$ is one of the ordered triples listed above, or all of $x,y,z$ are even and we can repeat this process. By the Well Ordering Principle, we cannot continue this forever, and so at some point, the ordered triple we get by dividing all the elements of the last ordered triple by $2$ must be one of the ones listed above.

Thus, we have that if $d=2e$ , then $(a,b,c)$ is a permutation of $(0,0,2^{e/2})$ and if $d=2e+1$ , then $(a,b,c)$ is a permutation of $(0,2^{e/2},2^{e/2})$ . Thus, each value of $d$ has exactly three possible ordered triples $(a,b,c)$ such that $a^2+b^2+c^2=2^d$ .

We have a total of $100$ possible values for $d$ , and so our answer is $3 \cdot 100 = \fbox{300}$ .

When d is 0 we find that the number of quadruples are 3. When d is 1 we find that the number of quadruples are also three. .... therefore as d <100 there are 3×100 quadruples