Find all the pairs of positive numbers such that the last digit of their sum is 3, their difference is a primer number and their product is a perfect square.
Problem
Source: Spanish Communities
Tags: LaTeX, modular arithmetic, quadratics, number theory, relatively prime, algebra, difference of squares
dyd
04.04.2006 12:43
suppose we seek for all pair $(a,b)$ where $a>b$. let $d=\gcd(a,b)$, then $a=dm$ and $b=dn$ where $\gcd(m,n)=1$. The condition becomes 1. $d(m+n)$ ends by 3 2. $d(m-n)$ is a prime 3. $mn$ is a square 4. from (3) and condition $\gcd(m,n)=1$, we have $m$ and $n$ are both squares. Let $m=e^2$ and $n=f^2$ 5. from (4), we conclude that $m-n\neq 1$ 6. from (5) and (2), we have $d=1$ and $m-n$ is a prime 7. from (6), we have $(e+f)(e-f)$ is a prime. It means that $e-f=1$ The rest is looking for $e,f$ such that $e+f$ is prime, $e-f=1$ , $e^2+f^2$ ends by 3. And it's very easy to verify.
paladin8
05.04.2006 08:31
We claim that the only such pair is $(4,9)$.
We want $x$ and $p$ such that $2x+p \equiv 3 \pmod{10}$ and $x(x+p) = y^2$. Consider $d = gcd(x,x+p)$. $d$ has to be $1$ or $p$.
If $d$ is $p$, then $\frac{x}{p}$ and $\frac{x}{p}+1$ have have product a perfect square but they are relatively prime so they are both squares. But we obviously don't have consecutive positive squares so there are no such cases.
If $d$ is $1$, then $x$ and $x+p$ are both squares. The difference of squares $p$ is a sum of consecutive odd integers, but then $p$ is divisible by the average of these integers and cannot be prime, except in the case where $p$ is the only consecutive odd integer. So $x = z^2$ and $x+p = (z+1)^2$.
So $x+(x+p) \equiv 3 \pmod{10}$ where $x$ and $x+p$ are consecutive quadratic residues. Looking at quadratic residues $\pmod{10}$, we find $0,1,4,9,6,5,6,9,4,1$. The only consecutive ones that sum to $3 \pmod{10}$ are $2^2+3^2$ and $7^2+8^2$.
So we have the case $p = (10k+3)^2-(10k+2)^2 = 20k+5 = 5(4k+1)$. This is prime iff $k = 0$ so $p = 5$ giving us $(4,9)$.
Also we have the case $p = (10k+8)^2-(10k+7)^2 = 20k+15 = 5(4k+3)$ which is never prime. So there are no solutions.
Hence the only pair is $(4,9)$ as claimed.
rafaello
14.12.2020 17:33
Let those numbers be $a,b$. Suppose $\gcd(a,b)=d>1$. Write $a=da_1$ and $b=db_1$, where $\gcd(a_1,b_1)=1$. $d(a_1-b_1)=p$ $d(a_1+b_1)\equiv 3\pmod{10}$ Since, $d>1$, we get that $a_1-b_1=1$ And thus, $$b_1(b_1+1)=x^2$$Since $b_1$ and $b_1+1$ are co-prime, we have that $b_1=y^2$ and $b_1+1=z^2$, where $y,z$ are positive. Hence, $(z-y)(z+y)=1$, since $y,z\geq 1$, we have $z+y\geq 2$ and since $z-y\in \mathbb Z$; a contradiction. Thus, $\gcd(a,b)=1$, therefore we have that $a=m^2$ and $b=n^2$, where $m,n$ are just positive integers. Let $a>b$, meaning $m>n$. Therefore, $m^2-n^2=p$. Factorising $(m-n)(m+n)=p$, thus $m+n=p$ and $m-n=1$ or solving it, we get that $2m=p+1$ and $2n=p-1$. Since their sum is $\equiv 3\pmod{10}$, we have that $\frac{p^2+1}{2}=\frac{2p^2+2}{4}=\frac{(p-1)^2+(p+1)^2}{4}=a+b\equiv 3\pmod{10}$. We also conclude that $\frac{p^2+1}{2}\equiv 3\pmod{10}$. By CRT, $\frac{p^2+1}{2}\equiv 1\pmod{2}$ and $\frac{p^2+1}{2}\equiv 3\pmod{5}$ Hence, $p^2\equiv 0\pmod{5}$, i.e. $p=5$. We conclude that $m=3$, $n=2$, thus $a=9$, $b=4$. Only pair is $(9,4)$.