An interesting problem in number theory
Solution 1
Hint: If $a^2+b^2=c^2$ then $(a+b)^2+(a-b)^2 = 2c^2$.
Solution 2
Thanks for your attention to this question. This question turns out to be a consequence of a well-known result in number theory (Fermat theorem on sums of two squares), see for instance, "Sally, Sally. Roots to Research. American Mathematical Society. 2007."
This results tells us, for any given non-negative integer $p$, how may possible ways can we solve $m^2+n^2=p$ in integer domain. Applying this result to the sequence equation $$ m^2+n^2=2*3^{2j}, \quad \quad j=0,1,2,\cdots\tag{3} $$
we see that (3) has 8 trivial solutions in integer domain (repeated), and so it has only the trivial solution $m=n=3^j$ in non-negative integer domain. Hence, $k=3^j\in K$ for each $j=0,1,2,\cdots$. This proves $K$ is unbounded.
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teh
Updated on November 19, 2020Comments
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teh almost 3 years
Thank you guys' contributions. I here reformulate this problem. For any given non-negative integer $k$, we consider the integer equation (meaning solving it in non-negative numbers) $$m^2+n^2=2k^2.\tag{1}$$
Clearly, this equation has a trivial solution $m=n=k$. Simple tests from $k=1$ to $k=13$ show that equation (1) has more than one solution for $k=5$, $k=10$ and $k=13$. More precisely, we have $$ 2\times 5^2=5^2+5^2=1^2+7^2, \quad 2\times 10^2=10^2+10^2=2^2+14^2, $$ and $$ 2\times 13^2=13^2+13^2=7^2+17^2. $$ In general, we have $$ 2\times (5k)^2=(5k)^2+(5k)^2=k^2+(7k)^2. $$ This problem is similar to Fermat's theorem on sums of two squares, right?. Yes, it is in the following sense (Thomas): If $k^2=a^2+b^2$ for some non-negative integers $a$ and $b$, then $$ 2k^2=k^2+k^2=(a-b)^2+(a+b)^2 $$ or write it as the form of (1): $$ 2(a^2+b^2)=(a^2+b^2)+(a^2+b^2)=(a-b)^2+(a+b)^2. $$ We see, for such number $k$, that equation (1) has at least two solutions. Further honest calculations: for k=185, equation (1) has solutions [[37, 259], [49, 257],[119, 233], [235, 115], [185, 185]] and for k=195 it has has [[141, 237], [273, 39], [69, 267], [255, 105], [195, 195]]
Based on these observations, we see this problem is not simple. Our interested problem is the following:
Weak conjecture : Does there exist a sub-sequence $\{k_i\}$ of natural numbers with $k_i\rightarrow \infty$ such that $$m_i^2+n_i^2=2k_i^2.\tag{2}$$ has a unique solution $m_i=n_i=k_i$? Equivalently, let $K$ denote the set of $k$ such that equation (1) has a unique solution $m=n=k$. Then the conjecture is that the set $K$ is unbounded in $\mathbb{N}$.
People also claimed that for any $N$ there is a $k$ such that (1) has at least $N$ solutions. But this result does not also disprove the above weak conjecture.
I hope this clarification will make this question more interesting. I personally think it may be a tough question. Looking forward to your great ideas!!!!
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Thomas Andrews about 10 yearsFor the weak conjecture, yes, if $k_i=3^{i}$ then the only solutions to $m_i^2+n_i^2=2k_i^2$ is of the above form. The first conjecture is false. Try $k=13$. Same for conjecture 2 with $k=13$.
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teh about 10 yearsThe second conjecture is false, since $2\times 13^2=13^2+13^2=7^2+17^2$. The issue now is the conjecture 3. I do not see why you claimed conjecture 1 is false?
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Thomas Andrews about 10 yearsThere are more than $2$ solutions to $m^2+n^2=2\cdot (65)^2$
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Thomas Andrews about 10 yearsThere's a general formula for the number of ways to write a natural number as the sum of two squares.
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Thomas Andrews about 10 yearsFor example: $89^2+23^2 = 79^2+47^2 = 85^2+35^2 = 2\cdot 65^2$. Indeed, for any $N$ we can find a value for $k$ for which there are at least $N$ solutions to $m^2+n^2=2k^2$.
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