Question

Find all non-negative integers a,b,c,d,n such that $a^2+b^2+c^2+d^2=7\cdot 4^n$.

• A. $(2^{n+1},2^n,2^n,2^n,2^n),(3\cdot 2^n,3\cdot 2^n,3\cdot 2^n)$
• B. $(3^{n-1},2^n,2^n,2^n,2^n),(3\cdot 2^n,3\cdot 2^n,3\cdot 2^n)$
• C. $(2^{n-1},2^n,2^n,2^n,2^n),(3\cdot 2^n,3\cdot 2^n,3\cdot 2^n)$
• D. $(3^{n+1},2^n,2^n,2^n,2^n),(3\cdot 2^n,3\cdot 2^n,3\cdot 2^n)$

Answer 1

The correct option is A $(2^{n+1},2^n,2^n,2^n,2^n),(3\cdot 2^n,3\cdot 2^n,3\cdot 2^n)$

For n=0, we have $2^2+1^2+1^2+1^2=7$, hence (a,b,c,d)=(2,1,1,1)and all permutations. If n $\ge$ 1, then $a^2+b^2+c^2+d^2\equiv 0\left(mod4\right)$, hence the numbers have the same parity,We analyze two cases.(a) The numbers a,b,c,d are odd. We write a = $2a^{,}+1,$ etc.We obtain $44^{\prime }(a^{\prime }+1)+4b^{\prime }(b^{\prime }+1)+4c^{\prime }(c^{\prime }+1)+4d^{\prime }(d^{\prime }+1)=4(7\cdot 4^{n-1}-1)$.The left-hand side of the equality is divisible by 8, hence $7\cdot 4^{n-1}-1$ must be even. This happens only for n=1. We obtain $a^2+b^2+c^2+d^2=28$, with the solutions (3,3,3,1) and (1,1,1,5).(b)The numbers a,b,c,d are even. Write $a=2a^{\prime },$ etc. We obtain $a^{\prime 2}+b^{\prime 2}+c^{\prime 2}+d^{\prime 2}=7\cdot 4^{n-1}$,so we proceed recursively.Finally ,we obtain the solutions $(2^{n+1},2^n,2^n,2^n,2^n),(3\cdot 2^n,3\cdot 2^n,3\cdot 2^n)$