A positive integer is called good if it can be written as the sum of two distinct integer squares. A positive integer is called better if it can be written in at least two was as the sum of two integer squares. A positive integer is called best if it can be written in at least four ways as the sum of two distinct integer squares. a) Prove that the product of two good numbers is good. b) Prove that $ 5$ is good, $ 2005$ is better, and $ 2005^2$ is best.
Problem
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Tags: number theory
eagles2018
17.09.2021 01:08
Let the good numbers be $a^2+b^2$ and $c^2+d^2$. Then $(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2=(ac+bd)^2+(ad-bc)^2$, which is a good number by definition. (and also a better number)
1) $5=1^2+2^2$
2) Using the above factorization, $2005=5\cdot 401=(1^2+2^2)(20^2+1^2)=(20-2)^2+(1+40)^2=18^2+41^2$ and $2005 = (1^2+2^2)(20^2+1^2)=(40-1)^2+(20+2)^2=39^2+22^2$. Thus, $2005=18^2+41^2=22^2+39^2$
3) We know that $2005^2=25 \cdot 40501=(3^2+4^2)(201^2+10^2)$, from which we can get $2$ sums using the factorization, and $2005^2=10025\cdot401=(100^2+5^2)(20^2+1^2)$, from which we can get two more sums using the factorization. Thus, $2005^2$ is best.
OlympusHero
17.09.2021 01:08
a) Let the two good numbers be $a^2+b^2$ and $c^2+d^2$. I claim that $(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2=(ac+bd)^2+(ad-bc)^2$. Indeed, expanding the LHS gives $a^2c^2+a^2d^2+b^2c^2+b^2d^2$ and expanding the RHS gives $a^2c^2+b^2d^2+a^2d^2+b^2c^2$, as desired.
b) Clearly $5=2^2+1^2$. Next, we have $2005=(2^2+1^2)(20^2+1^2)=39^2+22^2=41^2+18^2$. Now note that $(2mn)^2+(m^2-n^2)^2=(m^2+n^2)^2$, so $1476^2+1357^2=2005^2$ and $1716^2+1037^2=2005^2$. We can also scale up the $3-4-5$ triple, giving $1604^2+1203^2=2005^2$. Finally, we could scale up a triple with hypotenuse $401=20^2+1^2$. Using our previous identity gives $40^2+399^2=401^2$, so we have $200^2+1995^2=2005^2$, making a total of four ways as desired.