How to prove that the equation $3(a^2+b^2)=7(c^2+d^2)$ has no solution

Question

Question 1: Let $a,b,c,d$ be positive integers, show that

$$3(a^2+b^2)=7(c^2+d^2)$$ has no solution.

question 1 is from Mathlove, Curious How prove it?

My question 2: Find the least positive integer $k$ such that $$a^2+b^2=k(c^2+d^2)$$ has no solution.

Please put question 2 in a new question. – PM 2Ring Mar 21 '15 at 10:04 — PM 2Ring, Mar 21 '15 at 10:04

mathlove · Accepted Answer · 2015-03-21T10:03:15.543

1

HINT for Question 1 :

We have $$c^2+d^2\equiv 0\pmod 3\Rightarrow c\equiv d\equiv 0\pmod 3$$ and $$a^2+b^2\equiv 0\pmod 7\Rightarrow a\equiv b\equiv 0\pmod 7.$$

So, since we have $a=7p,b=7q,c=3r,d=3s,$ we have $$3(a^2+b^2)=7(c^2+d^2)\Rightarrow 7(p^2+q^2)=3(r^2+s^2).$$

edited Mar 21 '15 at 10:03

answered Mar 21 '15 at 09:56

mathlove

151,597

Thanks,But you think the question 2 find it? – Mar 21 '15 at 10:01
The analysis mod $7$ is unneeded, since we have $,9\mid$ RHS so $,3\mid a^2+b^2,\Rightarrow, 3\mid a,b,$ hence descent. – Bill Dubuque Mar 21 '15 at 16:34
@BillDubuque: You are right. Thank you for pointing it out. – mathlove Mar 21 '15 at 17:23

score 1 · Answer 2 · answered Mar 21 '15 at 09:58

1

$$\frac{a^2+b^2}{c^2+d^2}=\frac{(a^2+b^2)(c^2+d^2)}{(c^2+d^2)^2}$$

Use Brahmagupta–Fibonacci identity for the numerator to get $$\frac{a^2+b^2}{c^2+d^2}=\left(\frac{ac\pm bd}{c^2+d^2}\right)^2+\left(\frac{ad\mp bc}{c^2+d^2}\right)^2$$

But $3,7\equiv-1\pmod4$ can not be expressed as the sum of two squares

answered Mar 21 '15 at 09:58

lab bhattacharjee

279,016

Hello,so you think which is minumin of the $k$? – Mar 21 '15 at 10:01
@abandon, If $k>1,k=2$ as $2(a^2+b^2)=(a+b)^2+(a-b)^2$ – lab bhattacharjee Mar 21 '15 at 10:02
Hello,But $\dfrac{7}{3}\equiv \dfrac{3}{3}=1\pmod 4$? – Mar 21 '15 at 10:06
@abandon, We need each of $7,3$ to be of the form sum of two squares – lab bhattacharjee Mar 21 '15 at 10:10
why can't expressed sum of two squares? – Mar 21 '15 at 10:11
@abandon, http://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares – lab bhattacharjee Mar 21 '15 at 10:11
Thank you,I have use this methods to solve it,http://math.stackexchange.com/questions/1199216/show-that-fraca2b2c215-is-non-square-integer – Mar 21 '15 at 10:17
because $$a^2+b^2+c^2\equiv 0\pmod 8$$ – Mar 21 '15 at 10:17

score 1 · Answer 3 · answered Mar 21 '15 at 10:02

The famous two-square theorem by Fermat implies that $3n$ is not the sum of two squares, if $n$ is. This is answers Question $2$.

The first question is a nice example of infinite descent. One can show that all four numbers are dibisible by $3$, if the equation holds. Hence we get a smaller solution by dividing the equation by $9$.

JMP · Answer 4 · 2015-03-21T10:14:54.610

0

Q1: c and d are both $0\mod3$, so RHS is divisible by 9, which means a and b are both $0 \mod 3$, so LHS divisible by 27, etc...

Q2: k=3 by similar argument to Q1.

edited Mar 21 '15 at 10:14

answered Mar 21 '15 at 10:00

JMP

22,637

How to prove that the equation $3(a^2+b^2)=7(c^2+d^2)$ has no solution

4 Answers4