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Prove:

If $k \geq 3$, then $2^k$ can be written as $(2m+1)^2+7(2n+1)^2$, where $k, m, n \in \mathbb{N}$.

Benjamin Dickman
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Veritas
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  • First of all if $2^k$ divides $(2m+1)^2+7(2n+1)^2,$

    $$ (2m+1)^2\cdot{(2n+1)^{-1}}^2\equiv-7\pmod {2^k}$$

    As $-7\equiv1\pmod 8,$ we can prove (http://www.johndcook.com/quadratic_congruences.html) there are exactly, $4$ solutions if $k\ge 3$

    $\implies \exists m,n\in N,$ such $2^k$ divides $(2m+1)^2+7(2n+1)^2$

    Now using Brahmagupta-Fibonacci Identity (http://www.proofwiki.org/wiki/Brahmagupta-Fibonacci_Identity), $(a^2+7b^2)(c^2+7d^2)=(ac\pm 7bd)^2+7(ad\mp bc)^2$ We need to show each factor of $a^2+7b^2$ is of the same form

     – lab bhattacharjee May 26 '13 at 19:11
  • Does the tag "recrational-mathematics" really belong on this question? – Caleb Stanford May 27 '13 at 09:38
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    This question was proposed at the MMO 1985.It is due to Euler who never published it. – Shaswata Jun 08 '13 at 07:43

2 Answers2

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You need to be careful about just one thing in the induction step. Note that the first case is $$ 1^2 + 7 \cdot 1^2 = 8 = 2^3. $$ The important thing is that $$ 1 \equiv 1 \pmod 4. $$

So, we begin with $$ a^2 + 7 b^2 = 2^k $$ with $k \geq 3$ AND $$ a \equiv b \pmod 4, $$ then we do the induction step.

What the formula of Brahmagupta does for you is to allow us to multiply the value $2^k$ by 8, as in $$ (a - 7 b)^2 + 7 (a+b)^2 = 8 \cdot 2^k = 2^{k+3}. $$ At this stage, both numbers $a-7b,a+b$ are even, which is bad. On the other hand, $$ a - 7 b \equiv a+b \pmod 8. $$ As a result, when we divide by 2, we get $$ \frac{ a - 7 b}{2} \equiv \frac{a+b}{2} \pmod 4. $$ Furthermore, as $ a \equiv b \pmod 4, $ we know that $ a + b \equiv 2 \pmod 4, $ so that $\frac{a+b}{2} $ is once again ODD. We have completed the induction step, as $$ \left( \frac{ a - 7 b}{2} \right)^2 + 7 \left( \frac{a+b}{2} \right)^2 = 2 \cdot 2^k = 2^{k+1}, $$ with both numbers odd and the extra property we need to continue the induction,$$ \frac{ a - 7 b}{2} \equiv \frac{a+b}{2} \pmod 4. $$

It may be worth mentioning that we need to allow negative values for $(a,b)$ in order to keep the mod 4 thing going. So the pairs go $$ (1,1), \; \; (-3,1), \; \; (-5,-1), \; \; (1,-3), \; \; (11,-1), \; \; (9,5), $$ $$ (-13,7), \; \; (-31,-3), \; \; (-5,-17), \; \; (57,-11), \; \; (67,23), \; \; (-47,45), $$ $$ (-181,-1), \; \; (-87,-91), \; \; (275,-89), \; \; (449,93), \; \; (-101,271), \; \; (-999,85), \ldots $$

Will Jagy
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Hint: $2=(\frac{1+\sqrt{7}i}{2})(\frac{1-\sqrt{7i}}{2})$.

Ma Ming
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