Japan Mathematical Olympiad

Assume the existence of a positive integer solution $(a,b,c,d,n)$ such that $a^2+b^2+c^2+d^2-4\sqrt{abcd}=7\cdot 2^{2n-1}$, and take the one with the minimum value of $n$. Since $\sqrt{abcd}$ is rational, $abcd$ must be a square number. First, we establish the following two lemmas. Note that $a^2+b^2+c^2+d^2-4\sqrt{abcd}=(a-b{)}^2+(c-d{)}^2+2(\sqrt{ab}-\sqrt{cd}{)}^2$.

Lemma 1. $a,b,c,d$ are all distinct from each other. Proof. From the symmetry of $a,b,c,d$, it suffices to assume $a=b$ and derive a contradiction. Since $abcd=a^{2}cd$ is a square number, $cd$ must also be a square number. Therefore, both $M=c-d$ and $N=a-\sqrt{cd}$ are integers, and we have $M^2+2N^2=7\cdot 2^{2n-1}$. Since the residues of square numbers divided by $7$ are $0,1,2,$ or $4$, the residues of $M^2$ and $2N^2$ when divided by $7$ must also be $0,1,2,$ or $4$. Since their sum is a multiple of $7$, both $M$ and $N$ must be multiples of $7$. Therefore, $M^2+2N^2$ is a multiple of $49$. However, $7\cdot 2^{2n-1}$ is not a multiple of $49$, which is a contradiction. ■

Lemma 2. The product of any two numbers among $a,b,c,d$ is different from the product of the remaining two numbers. Proof. From the symmetry of $a,b,c,d$, we only need to derive a contradiction by assuming $ab=cd$. Let $M=a-b$ and $N=c-d$, then $M$ and $N$ are integers, and $M^2+N^2=7\cdot 2^{2n-1}$ holds. Because the residues of square numbers divided by $7$ can only be $0,1,2,$ or $4$, as mentioned in Lemma 1, both $M$ and $N$ must be multiples of $7$. Therefore, $M^2+N^2$ is a multiple of $49$, but $7\cdot 2^{2n-1}$ is not, leading to a contradiction. ■

Consider cases based on the value of $n$.

(1) $n=1$ By the symmetry of $a,b,c,d$ and Lemma 1, we can assume that $a<b<c<d$. From $(d-a{)}^2+(c-b{)}^2+2(\sqrt{ad}-\sqrt{bc}{)}^2=14$, we have $d-a\le \sqrt{14}<4$, implying $a=b-1=c-2=d-3$. We also get $bc=a^2+3a+2=ad+2$ and consequently $\sqrt{ad}+2-\sqrt{ad}=\sqrt{2}$, which leads to a contradiction since $a$ and $d$ are positive.

(2) $n>1$ When $a,b,c,d$ are all even, $(\frac{a}{2},\frac{b}{2},\frac{c}{2},\frac{d}{2},n-1)$ satisfies the given equation, contradicting the minimality of $n$. Therefore, at least one of $a,b,c,d$ is odd. Since the residues of square numbers divided by $4$ are $0$ or $1$, and $a^2+b^2+c^2+d^2=7\cdot 2^{2n-1}+4\sqrt{abcd}$ is divisible by $4$, all of $a,b,c,d$ are odd. Since $abcd$ is an odd square number, $abcd\equiv 1(\mathrm{mod}4)$, implying that there are an even number of elements among $a,b,c,d$ that leave a remainder of $3$ when divided by $4$. Thus, without loss of generality, we can assume $a\equiv b(\mathrm{mod}4)$ and $c\equiv d(\mathrm{mod}4)$. Let $g$ denote the greatest common divisor of $ab$ and $cd$. Then, $\frac{ab}{g}$ and $\frac{cd}{g}$ are coprime, and $\frac{ab}{g}\cdot \frac{cd}{g}=\frac{abcd}{g^2}$ is a square number. Hence, there exist positive integers $x$ and $y$ such that $ab=g{x}^2$ and $cd=g{y}^2$, and $(\sqrt{ab}-\sqrt{cd}{)}^2=g(x-y{)}^2$ holds. Since $a\equiv b(\mathrm{mod}4)$, we have $ab\equiv 1(\mathrm{mod}4)$, and combined with $x^2\equiv 1(\mathrm{mod}4)$, this yields $g\equiv 1(\mathrm{mod}4)$. By Lemma 1 and Lemma 2, we can find positive odd integers $p,q,r$ and non-negative integers $\alpha ,\beta ,\gamma$ satisfying $|a-b|=p\cdot 2^{\alpha },|c-d|=q\cdot 2^{\beta }$, and $|x-y|=r\cdot 2^{\gamma }$. Remark that $$\begin{array}{rl}7\cdot 2^{2n-1}& =a^2+b^2+c^2+d^2-4\sqrt{abcd}\\ & =(a-b{)}^2+(c-d{)}^2+2(\sqrt{ab}-\sqrt{cd}{)}^2\\ & =p^2\cdot 2^{2\alpha }+q^2\cdot 2^{2\beta }+g{r}^2\cdot 2^{2\gamma +1}\end{array}$$ holds. $p^2\equiv q^2\equiv 1(\mathrm{mod}8)$, $g{r}^2\equiv 1(\mathrm{mod}4)$ imply a contradiction by the following lemma.

Lemma 3. There exist no non-negative integers $\alpha ,\beta ,\gamma$, positive integers $n,s,t,u$ such that $s\equiv t\equiv 1(\mathrm{mod}8)$, $u\equiv 1(\mathrm{mod}4)$, and $$s\cdot 2^{2\alpha }+t\cdot 2^{2\beta }+u\cdot 2^{2\gamma +1}=7\cdot 2^{2n-1}.$$ Proof. By the symmetry of $(\alpha ,s)$ and $(\beta ,t)$, we can assume $\alpha \le \beta$.

case when $\alpha \le \gamma$ Since $s+t\cdot 2^{2(\beta -\alpha )}+u\cdot 2^{2(\gamma -\alpha )+1}=7\cdot 2^{2(n-\alpha )-1}$ is a positive integer, we must have $n-\alpha \ge 1$. Therefore, both $u\cdot 2^{2(\gamma -\alpha )+1}$ and $7\cdot 2^{2(n-\alpha )-1}$ are even, implying that $s+t\cdot 2^{2(\beta -\alpha )}$ is even as well, and consequently, $\alpha =\beta$. Since $s+t\cdot 2^0$ cannot be divisible by $4$, we obtain either $\gamma -\alpha =0$ or $n-\alpha =1$. In the former case, $7\cdot 2^{2(n-\alpha )-1}=s+t+2u\equiv 4(\mathrm{mod}8)$, leading to a contradiction. In the latter case, $u\cdot 2^{2(\gamma -\alpha )+1}=14-s-t\equiv 4(\mathrm{mod}8)$, which also contradicts the conditions.

case when $\gamma <\alpha$ Since $s\cdot 2^{2(\alpha -\gamma )}+t\cdot 2^{2(\beta -\gamma )}+2u=7\cdot 2^{2(n-\gamma )-1}$ has a remainder of $2$ when divided by $4$, we must have $n-\gamma =1$. In this case, $2^{2(\beta -\gamma )}<14$, implying $\alpha -\gamma \le \beta -\gamma \le 1$, which leads to $\alpha -\gamma =\beta -\gamma =1$. Therefore, we have $2s+2t+u=7$, but $2s+2t+u\equiv 1(\mathrm{mod}4)$, which is a contradiction.

Therefore we have proved the lemma.

Therefore, there are no solutions $(a,b,c,d,n)$.