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Let a and b be a positive integers. Proof that if number $ 100ab -1 $ divides number $ (100a^2 -1)^2 $ then also divides number $ (100b^2 -1)^2 $.

My attempt:

Let's notice that \begin{split} b^2(100a^2-1)^2 -a^2(100b^2-1)^2 & =(100a^2b-b)^2-(100ab^2-a)^2\\ & =(100a^2b-b-100ab^2+a)(100a^2b-b+100ab^2-a)\\ & =(100ab(a-b)+(a-b))(100ab(a+b)-(a+b))\\ & =(a-b)(100ab+1)(a+b)(100ab-1).\end{split}

This means that $ 100ab-1 |a^2(100b^2-1)^2$, so if we proof that $\gcd(100ab-1,a^2)=1$ the proof is completed. Now I know that it should be trivial to show that these numbers are relatively prime but somehow i have no idea how to do it.

Also I am intrested if there is a way to solve this problem by using modular arithmetic?

Alice211
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2 Answers2

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For a modular arithmetic argument:

Let $N=100ab-1$ and work modulo $N$.

Starting with $(100a^2 -1)^2\equiv 0$, multiply by $10^4b^4$: $$(10^4a^2b^2-100b^2)^2\equiv 0.$$

Since $10^2ab\equiv 1$, we have $(1-100b^2)^2\equiv 0.$

  • Could you explain last line a bit more? I get that $10^2ab \equiv 1$ by why these means that $(1-100b^2)^2\equiv 0$. I am sorry if this is a stupid question by I am just starting to learn some basic number theory. – Alice211 Feb 03 '22 at 23:44
  • In the previous line I have replaced $10^4a^2b^2=(100ab)^2$ by $1$. –  Feb 03 '22 at 23:47
  • I think i get. So let's say that $(x-y)^2\equiv 0 $ (mod n)and we know that $x\equiv 1 $then we can write that $(1-y)^2\equiv 0$? – Alice211 Feb 03 '22 at 23:51
  • Yes that's right. Modular arithmetic arguments are so easy! –  Feb 03 '22 at 23:52
  • Thank you! I have only one question. What properties of modular arithmetic we are using when we replace x by 1. I mean it obvious for me that if $x\equiv 1$ then $x-y \equiv1-y$ but why this is also true when we have $(x-y)^2$? Is it because we can multiply congruence's? – Alice211 Feb 04 '22 at 00:03
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    Since $x\equiv 1$, we have $x-y =1+kn-y$ for some $k$. When this is squared we can ignore all multiples of $N$ if we are working modulo $N$. –  Feb 04 '22 at 00:07
  • When carrying out such a modular arithmetic argument just treat $x$ and $1$ as the same. –  Feb 04 '22 at 00:10
  • Okay. Thank You once again! – Alice211 Feb 04 '22 at 00:12
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    @S.Dolan Careful. It's not always the case, when dealing with modular arithmetic, that a number $x$ can be replaced by any other number $y$ congruent with $x$. – jjagmath Feb 04 '22 at 00:50
  • It is always the case for the elements of the ring of integers mod N involved in the arithmetical operations. It is not, of course, the case for other elements such as those used in the standard notation for powers. –  Feb 04 '22 at 09:17
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    @Alice211 $,x\equiv a,\Rightarrow,f(x)\equiv f(a),$ for any polynomial $,f(x),$ with integer coef's - see the Polynomial Congruence Rule and other basic congruence arithmetic laws proved there. I added an answer which gives a more motivated derivation using modular arithmetic, which shows how it is a special case of a more general well-known method. – Bill Dubuque Feb 04 '22 at 11:53
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We show how modular arithmetic allows us to view it as a special case of polynomial reversal, i.e. that $f(a)=0\Rightarrow \tilde f(a^{-1})=0,\,$ where $\tilde f$ denotes the reverse (reciprocol) polynomial.

Here mod arithmetic works nicely: $\bmod 100ab-1\,$ we have $\,100ab\equiv 1\,$ so $\,\color{#c00}{a \equiv 1/(100b)}.\,$ We can substitute this into any polynomial equation $\,f(\color{#c00}a)\equiv 0\,$ then clear denom's to get an equation $\,g(b)\equiv 0\,$ for $\,b.\,$ Here $f(a)\equiv (100\color{#c00}a^2-1)^2\equiv0\,$ so making said $\rm\color{#c00}{substitution}$ for $\,\color{#c00}a\,$ yields

$$0\equiv f(a) \equiv \left[\dfrac{100}{(\color{#c00}{100b})^2}-1\right]^2 \equiv \left[\dfrac{1-100b^2}{\color{#0a0}{100b^2}}\right]^2\!\Rightarrow (1-100b^2)^2\equiv 0\!\!$$

Your proof, viewed modularly, essentially squares the following equation (compare here)

$$\begin{align} b(\color{#c00}{100a}a-1) &\,\equiv\, a(1-\color{#0a0}{100b}b)\\ \iff\ \ b\ (\color{#c00}{b^{-1}}\ a-1) &\,\equiv\, a(1-\ \color{#0a0}{a^{-1}}\ b),\,\ \text{is true by both} \equiv a-b \end{align}\qquad$$

So squaring we get $\,(100aa-1)^2\equiv 0\Rightarrow \color{#0a0}{a^2}(1-100bb)^2\equiv 0\Rightarrow (1-100bb)^2\equiv 0\,$ by twice cancelling the unit (invertible) $\,\color{#0a0}a\,$, i.e. by scaling by $\,\color{#0a0}{a^{-1}\, (\equiv 100b)}$.

Remark $ $ We can do all arithmetic fraction-free by scaling $\,f(a)\,$ by $\,(\color{#0a0}{100b^2})^2$ (this is essentially what is done in S. Dolan's answer, but there the key idea $\rm\color{#c00}{(elimination)}$ is not brought to the fore).

Above is a slight variation of the following well known result: $ $ if $\,a\,$ is a root of a polynomial $\,f(x)\,$ then $\,a^{-1}\,$ is a root of the reciprocal (reverse) polynomial $\,x^{d}f(1/x),\,$ $\, d := \deg f,\,$ as above.

Thus by using modular arithmetic we can express the problem using equations (congruences) and this allows us to use well-known facts on the relationship between an equation for $\,a\,$ and one for its inverse $\,a^{-1}.\,$ This relationship would be obfuscated if we used only divisibility language (vs. congruence equations).

Bill Dubuque
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