Prove that for all positive integers n there exist n distinct, positive rational numbers with sum of their squares equal to n. Proposed by Daniyar Aubekerov
Problem
Source:
Tags: number theory
Puzzled417
15.01.2017 07:32
We first prove that there exist infinitely many pairs of rational numbers (x,y) such that x2+y2=2.Take any Pythagorean triple (a,b,c) with b>a (note that there are infinitely many such Pythagorean triples). Then we can take x=b−ac,y=b+ac, which gives x2+y2=2.
Now take any positive integer n. If n is even, then take n2 pairs of rationals with sum of squares equal to 2. If n is odd, then take n−12 such pairs, and one number 1.
We now prove that all the rationals are distinct. If b+ac=b−ac, then a=0, an impossibility. Thus we consider two different primitive Pythagorean triples, (a,b,c) and (a′,b′,c′) (with c<c′) that generate at least two of the same rational numbers.
Since the sum of squares of those rationals are equal to 2, the other pair of rationals must be equal. Thus we have either b−ac=b′−a′c′andb+ac=b′+a′c′⟹b−ab′−a′=b+ab′+a′=cc′=λ∈Q, orb−ac=b′+a′candb+ac=b′−a′c′⟹b−ab′+a′=b+ab′−a′=cc′=λ∈Q.In both cases we have a2+b2=c2=λ2(c′)2=λ2((a′)2+(b′)2) and b2−a2=λ2((b′)2−(a′)2). Thus c2=λ2(c′)2,a2=λ2(a′)2,b2=λ2(b′)2. But then since λ=pq where gcd, q \mid a',b',c', which contradicts the fact that our triples are primitive. Thus all our rationals are distinct.
iman007
16.12.2020 23:33
Puzzled417 wrote:
We first prove that there exist infinitely many pairs of rational numbers (x,y) such that x^2+y^2 = 2.Take any Pythagorean triple (a,b,c) with b > a (note that there are infinitely many such Pythagorean triples). Then we can take x = \dfrac{b-a}{c},y = \dfrac{b+a}{c}, which gives x^2+y^2 = 2.
Now take any positive integer n. If n is even, then take \dfrac{n}{2} pairs of rationals with sum of squares equal to 2. If n is odd, then take \dfrac{n-1}{2} such pairs, and one number 1.
We now prove that all the rationals are distinct. If \dfrac{b+a}{c} = \dfrac{b-a}{c}, then a = 0, an impossibility. Thus we consider two different primitive Pythagorean triples, (a,b,c) and (a',b',c') (with c < c') that generate at least two of the same rational numbers.
Since the sum of squares of those rationals are equal to 2, the other pair of rationals must be equal. Thus we have either \begin{align*}\dfrac{b-a}{c} &= \dfrac{b'-a'}{c'} \quad \text{and} \quad \dfrac{b+a}{c} = \dfrac{b'+a'}{c'} \implies \dfrac{b-a}{b'-a'} = \dfrac{b+a}{b'+a'} = \dfrac{c}{c'} = \lambda \in \mathbb{Q}, \text{ }\text{or}\\\dfrac{b-a}{c} &= \dfrac{b'+a'}{c} \quad \text{and} \quad \dfrac{b+a}{c} = \dfrac{b'-a'}{c'} \implies \dfrac{b-a}{b'+a'} = \dfrac{b+a}{b'-a'} =\dfrac{c}{c'} = \lambda \in \mathbb{Q}.\end{align*}In both cases we have a^2+b^2 = c^2 = \lambda^2(c')^2 = \lambda^2((a')^2+(b')^2) and b^2-a^2 = \lambda^2((b')^2-(a')^2). Thus c^2 = \lambda^2(c')^2,a^2 = \lambda^2(a')^2,b^2 = \lambda^2(b')^2. But then since \lambda = \dfrac{p}{q} where \gcd(p,q) = 1, q \mid a',b',c', which contradicts the fact that our triples are primitive. Thus all our rationals are distinct.
We first prove that there exist infinitely many pairs of rational numbers (x,y) such thatx^2+y^2 = 2.
you can easily consider a circle with diameter equal to \sqrt{2} then consider AB as its diameter since there are infinitely many points like P on the circle such that PA^2+PB^2=AB^2=2 so we are done for this part