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Q: What is the value of $ Let $a$ and $b$ be coprime integers. Prove that $a \equiv b \pmod{(a-b)^2}$. I have rewritten this as follows: Let $a$ and $b$ be coprime integers. Prove that $a \equiv b \pmod{(a-b)}$. Are both of these equivalent? If so, how? A: $$(a-b)^2=a^2-2ab+b^2=(a+b)(a-b)=a-b+2ab$$ thus $$a\equiv a\bmod(a-b)^2\iff a\equiv a\bmod 2ab\iff a\equiv a\bmod ab$$ A: There are two different ways, but it is always good to go two ways. You can consider it modulo $a-b$ and modulo $a+b$. If you do it modulo $a-b$, then all the divisors of $a-b$ are also divisors of $a+b$, so $a\equiv b\mod (a-b)^2$ if and only if $a\equiv b \mod (a-b)$ If you do it modulo $a+b$ then, with the help of modulo 2, we have that the divisors of $a$ are also divisors of $a+b$, so $a\equiv b\mod(a+b)^2$ if and only if $a\equiv b\mod 2(a+b)$ So, if the result holds modulo $a+b$ and modulo $a-b$ the result holds. For the converse, we have that the 595f342e71


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