How to solve $at + b = 0 \pmod {(a-t)}$?

Question

Is there any way (except trying for $t=0,1,2,\ldots,a-1)$ to solve the following equation for $t$ when $a$ and $b$ are known?

$$ at +b = 0 \pmod{(a-t)} \text{ with } a,b,t \in N $$

Do you want to find one solution? Then take $t = a-1$, or do you want to find all? — martini, May 22 '13 at 18:47

M.H · Answer 1 · 2013-05-22T19:00:34.133

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$at +b = 0 \mod\ (a-t)$ then $at+b=-a^2+at \mod\ (a-t)$ we have $$b+a^2=0 \mod\ (a-t)$$ and $$b+a^2=k(t-a)$$ $$t=\frac{b}{k}+\frac{a^2}{k}+a: k\in \mathbb Zk|b ,k|a^2$$

edited May 22 '13 at 19:00

answered May 22 '13 at 18:54

M.H

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How did you get the first implication? Did you just put $-a(a-t)$? – Igor Deruga May 22 '13 at 19:02
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@Igor:yes if $a=b \mod p$ then we have $a=b+mk \mod m$ – M.H May 22 '13 at 19:07
How do you get $k|b ,k|a^2$? Doesn't it rather have to be $k|(b+a^2)$? – vauge May 23 '13 at 07:03
Nice Maisam. +1 – Mikasa May 23 '13 at 19:13
@Babak S.:thanks – M.H May 23 '13 at 19:14
Still, could someone answer my question? How do you get $k|b,k|a^2$, and shouldn't it be $k|(b+a^2)$ instead? – vauge May 24 '13 at 14:49

N. S. · Answer 2 · 2013-05-24T16:24:21.280

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$$a*t +b = 0 \mod (a-t) \Leftrightarrow at +b = k(a-t)$$ for some integer $k$.

This is equivalent to

$$at+kt=ka-b \Leftrightarrow t=\frac{ka-b}{a+k}=a-\frac{a^2+b}{a+k}$$

This is an integer if and only if $a+k$ is a divisor of $a^2+b$.

Thus,

Step 1: List all the integer divisors of $a^2+b$.

Step 2: For each divisor $d$ set $k=d-a$ and then $t=a-\frac{a^2+b}{d}$. Keep only the positive solutions.

edited May 24 '13 at 16:24

answered May 22 '13 at 20:08

N. S.

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Shouldn't you actually get $at+kt=ka-b$ instead of $at+kt=ka+b$? – vauge May 24 '13 at 14:50
@valgel Fixed. I think... – N. S. May 24 '13 at 16:24

How to solve $at + b = 0 \pmod {(a-t)}$?

2 Answers2