Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $a\ne b$ be two positive integers. Are $4ab+1$ and $(4a^2+1)^2$ coprime always? Can you find $a$ and $b$ such that they are not coprime?


It has been proved that $4ab-1$ is not a divisor of $(4a^2-1)^2$. Are $4ab-1$ and $(4a^2-1)^2$ always coprime?

share|improve this question

4 Answers 4

up vote 6 down vote accepted

Hint $\ $ Counterexamples abound: choose a modulus $\rm\,m\,$ so that $\rm\:mod\ m\!:\ a\equiv b\:$ and $\rm\:4a^2\pm1\equiv 0.\:$ Then $\rm\:4ab\pm1\equiv 4a^2\pm1\equiv 0,\:$ hence $\rm\,m\,|\,4ab\pm1,4a^2\pm1.\:$

E.g. for any $\rm\:a,\,$ let $\rm\,m>1\,$ be a divisor of $\rm\,4a^2\pm1,\,$ $\rm\,b = a\!+\!m,\:$ e.g. $\rm\:a=1,\: m=4\!\pm\!1,\: b = 5\!\pm\!1.$

Remark $\ \ $ Perhaps it will prove a bit instructive to present how I derived the counterexamples. This will yield a precise criterion for coprimality.

$$\rm\begin{eqnarray}\rm (4a^2\!+\!1,\,4ab\!+\!1) &=&\rm (4a^2\!+\!1,\,4a(b\!-\!a)\!-\!(4a^2\!+\!1))\\ &=&\rm (4a^2\!+\!1,\,4a(b\!-\!a))\\ &=&\rm (4a^2\!+\!1,\,b\!-\!a)\ \ via\ \ \ (4a^2\!+\!1,4a) = 1\ \text{ and Euclid's Lemma}\end{eqnarray}$$

This implies the following general criterion

$$\rm (4a^2\!+\!1,(4ab\!+\!1)^2) = 1\ \iff\ (4a^2\!+\!1,\,b\!-\!a) = 1$$

Hence the only counterexamples arise as above: $\rm\ \ 1 < m\,|\,4a^2\!+\!1,\,b\!-\!a$.

Alternatively, one could employ the follow Brahmagupta sum of squares identity

$$\rm (1+4a^2)(1+4b^2)\ =\ (1+4ab)^2 + 4\, (a-b)^2 $$

which should lead to a nice viewpoint in terms of Gaussian integer arithmetic.

share|improve this answer
Are these all possible counterexamples? –  user59671 Feb 7 '13 at 21:56
@CutieKrait Yes, see the appended Remark. –  Math Gems Feb 7 '13 at 22:15

When $a=1$ and $b=6$, $4ab+1=25$ and $(4a^2+1)^2=25$.

When $a=1$ and $b=4$, $4ab-1=15$ and $(4a^2-1)^2=9$.

share|improve this answer
These are special cases of counterexamples that exist for all $\rm\,a.\:$ See my answer. –  Math Gems Feb 7 '13 at 21:41

When $b = 4a^3 + 2a$, $4ab+1$ is exactly equal to $(4a^2+1)^2$. Why would you think that the former could not divide the latter?

On the other hand, it is true that $4ab - 1$ cannot divide $(4a^2+1)^2$. This is because the latter is expressible as the sum of two relatively prime squares, and no positive integer congruent to $3 \pmod 4$ divides such a number.

share|improve this answer
It's interesting that $4ab-1$ does not divide both $(4a^2-1)^2$ and $(4a^2+1)^2$. thanks for last note. –  user59671 Feb 7 '13 at 21:43

Let me prove why $4ab−1$ is not a divisor of $(4a^2−1)^2$ when $a\ne b$. This proof is not mine and for me it is very interesting (it is a famous proof). Let $$A=\{(x,y) \mid x\ne y,\quad 4xy−1| (4x^2−1)^2\}$$ Fisrtly we show that $A$ is a symmetric relation on $\mathbb{N}$, that is $$(\forall (x,y)\in A)((y,x)\in A)\quad\quad\quad\quad(1)$$

Let $(x,y)\in A$ be arbitrary. $mod \space 4xy-1$ we have $$(4xy)^2 \equiv1$$ and so $$4y^2-1\equiv4y^2-(4xy)^2=-4y^2(4x^2-1)\equiv0$$ so $$4xy-1|4y^2-1$$ that is $(y,x)\in A$.

Next we show that: $$(\forall (x,y)\in A)(\exists (a,b)\in A)(ab<xy)\quad\quad\quad\quad(2)$$ Let $(x,y)\in A$ be arbitrary. Because $A$ is symmetric, without loss of generality we can assume $x<y$. Let: $$k:=\frac{(4x^2−1)^2}{4xy−1}<\frac{(4x^2−1)^2}{4x^2−1}=4x^2−1$$

$mod\space 4x$ we have $$ -1 \equiv 4xy-1$$ $$ \to -k \equiv k(4xy-1)=(4x^2−1)^2 \equiv 1$$ $$\to k\equiv-1$$ $$\to k = 4xz -1<4x^2−1$$

we have $$4xz -1|4x^2−1,\quad z<x<y$$ $$\to (x,z) \in A ,\quad xz<xy$$

So $(2)$ is proved.

Now it is clear that $A$ must be empty.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.