2
$\begingroup$

I am trying to solve the following problem.

Find perfect square which last 4 digit is 9009

The solution in the textbook starts as follows.

Let $x$ be the one we want to obtain. Then $x^2 = 10000y + 9009$. Then $x=10a \pm 3$.

I am confused with the last part. why $x=10a\pm 3$?

$\endgroup$
5
  • 1
    $\begingroup$ Since the square ends in $9$, $x$ must end in either $3$ or $7$. $\endgroup$
    – lulu
    Commented Jan 10, 2020 at 15:13
  • $\begingroup$ @lulu how one can prove that? $\endgroup$
    – phy_math
    Commented Jan 10, 2020 at 15:15
  • 2
    $\begingroup$ Trial and error. There are only $10$ digits $x$ might end in. Try all of them. $\endgroup$
    – lulu
    Commented Jan 10, 2020 at 15:16
  • $\begingroup$ @lulu thanks, I will try $\endgroup$
    – phy_math
    Commented Jan 10, 2020 at 15:17
  • $\begingroup$ @phy_math we can consider all possiblities of the ending digit from 0 to 9 $\endgroup$ Commented Jan 10, 2020 at 15:17

3 Answers 3

2
$\begingroup$

For any integer $M$ there are integers $a,b$ so that $M =10a + b$ and $b = 0....9$.

If $b >5$ then $M = 10(a+1) - (10-b)$ where $10-b = 4,3,2,1$.

So for any integer $M$ there are integers $c,d$ so that $M = 10c \pm d$ and $d= 0,1,2,3,4,5$

That means $M^2 = (10c \pm d)^2 = 100c^2 \pm 20cd + d^2 = 10(10c^2 \pm 2cd) + d^2$

Now $d= 0,1,2,3,4,5$ so $d^2 = 0,1,4,9,16,25$.

So if the last digit of $M^2$ ends with $0,1,4,9,6$ or $5$ then $M= 10c \pm 0,1,2,3,4$ or $5$ respectively. So if $M^2 = 10000y + 9009$ then $M = 10c \pm 3$.

So we have $M^2 = (10c \pm 3)^2 = 100c^2 \pm 60 c + 9= 10000y + 9009$

So $100c^2 \pm 60c = 10000y + 9000$. we divide both sides by $10$ we get

$10c^2 \pm 6c = 1000y + 900$. The right hand side is divisible by $10$ we which means $6c$ must also be divisible by $10$. That means $c$ is divisible by $5$.

So if we say $c = 5e$ then we have $10*(25e^2) \pm 30e = 1000y + 900$ and we divide both sides by $10$ and get

$25e^2 \pm 3e = 100y + 90$. Now the RHS is divisible by $5$ so that means $3e$ is divisible by $5$ which means $e$ is divisible by $5$. Let $e = 5f$. So we have

$25*25f^2 \pm 15f = 100y + 90$. Divide both sides by $5$ and we get

$125f^2 \pm 3f = 20y+18$

Now if $f$ is even then $125f^2$ will be divisible by $20$. And if $f = 6$ then $3f = 18$. So that will be one possible answer.

If $f = 6$ then $e= 5*f= 30$ and $c=5e = 150$. And $M=10c + 3 = 1503$.

And so $1503^3 = (1500^2 + 2*1500*3 + 3^2) = 2250000 + 9000 + 9 = 2259009$.

$\endgroup$
0
$\begingroup$

Let $x=10a\pm b,0\le b\le5$

$\implies b^2\equiv0,1,4,9,5\pmod{10}$

$x^2=10(10a^2\pm2ab)+b^2\equiv b^2\pmod{10}$

$\endgroup$
0
$\begingroup$

\begin{eqnarray*} 3497^2=12229009. \end{eqnarray*} There could be others.

$\endgroup$
1
  • $\begingroup$ actually $3487^2 = 12159169$ $\endgroup$
    – phy_math
    Commented Jan 11, 2020 at 5:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .