7
$\begingroup$

Find the smallest positive integer which can be written as the sum of the squares of two positive integers in two different ways.

I took extremely long to solve this

I got

$50= 7^2 + 1^2 $

$50= 5^2 + 5 ^2 $

I did it with the trial and error method.

I'm just curious if this came out during my exams, is there a quicker way to find the answer?

$\endgroup$
  • $\begingroup$ It is the variant of so called "Taxicab numbers", for better known 3rd and 4th power's variant an algorithm is described in Nikolaus Wirth's book Algorithmen und Datenstrukturen from 1975. $\endgroup$ – z100 Jun 20 '16 at 14:50
  • $\begingroup$ See oeis.org/A007692 and mathworld.wolfram.com/SumofSquaresFunction.html. $\endgroup$ – lhf Jun 20 '16 at 15:50
5
$\begingroup$

Consider the equation $x^2+y^2=z^2+w^2=N.$

This is equivalent to: $x^2-z^2=w^2-y^2=D.$

and we need $D$ to have at least two different factorizations with the factors having the same parity. One approach is thus to find the smallest values of $D$ which satisfy this property. Another approach is to find parametric solutions.

Let $(x-z)=ab$ and $(x+z)=cd$ with $(w-y)=ac$ and $(w+y)=bd$,

which gives the parametric solution: $\dfrac{1}{2}(ab+cd,bd-ac,cd-ab,ac+bd)$. We have $N=\dfrac{1}{4}(a^2b^2+c^2d^2+b^2d^2+a^2c^2)$.

Now from the fact that $cd-ab$ and $bd-ac$ are distinct positive integers, we can see that no three of $a,b,c,d$ can be the same.

Thus, the lexicographically smallest value for the tuple $(a,b,c,d)$ is $(1,1,2,3)$, which gives the solution $(7,1,5,5)$ and $N=50$. Once we have this solution, it is easily checked that other tuples of $a,b,c,d$ give larger values of $N$.

$\endgroup$
4
$\begingroup$

I do wonder why it took so long... I added all pairs from the first ten squares on a piece of scrap paper in a few minutes. Maybe that's a long time.

\begin{array}{c|c|c} && +1 & +4 & +9 & +16 & +25 & +36 & +49 & +64 & +81 & +100 \\ \hline 1 && 2 \\ \hline 4 && 5 & 8\\ \hline 9 && 10 & 13 & 18 \\ \hline 16 && 17 & 20 & 25 & 32 \\ \hline 25 && 26 & 29 & 34 & 41 & \color{red}{50} \\ \hline 36 && 37 & 40 & 45 & 52 & 61 & 72 \\ \hline 49 && \color{red}{50} & 53 & 58 & \color{blue}{65} & 74 & \color{magenta}{85} & 98 \\ \hline 64 && \color{blue}{65} & 68 & 73 & 80 & 89 & 100 & 113 & 128 \\ \hline 81 && 82 & \color{magenta}{85} & 90 & 97 & 106 & 117 & 130 & 145 & 162 \\ \hline 100 && 101 & 104 & 109 & 116 & 125 & 136 & 149 & 164 & 181 & 200 \\ \hline \end{array}

This also gives the first case when all the squares in question are distinct, at $65 = 8^2+1^2 = 7^2+4^2$

$\endgroup$
0
$\begingroup$

If $p$ is a prime and $p \equiv 1 \bmod 4$, then $p=a^2+b^2$, for some $a,b \in \mathbb N$.

In this case, $2p^2$ has two representations: $p^2+p^2$ and $(a-b)^2+(a+b)^2$.

The second representation ones from $2p^2=|(1+i)(a+bi)^2|^2$.

So, the smallest solution is with $p=5$, for which $a=2$ and $b=1$ and $(a+bi)^2=3-4i$, which give $50=5^2+5^2=1^2+7^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.