Can $\sqrt{ab}$ and $\sqrt{a^2 + b^2}$ be both integers if $a$ and $b$ are natural numbers?

Question

Does there exist an $a \in \mathbb{N}$ and $b \in \mathbb{N}$ such that $\sqrt{ab} \in \mathbb{Z}$ and $\sqrt{a^2 + b^2} \in \mathbb{Z}$?

Answer 1

no, but it is pretty elaborate. If there is a solution with positive integers, it is still a solution if we divide out by the greatest common divisor of $a,b.$ We may therefore demand $\gcd(a,b) = 1.$

If $ab$ is a square and they are coprime, both $a,b$ are squares.

You are asking for $a = m^2, b = n^2,$ and you want $m^4 + n^4 = c^2.$ This has no nonzero solutions, goes back to Fermat. In fact, this is the first thing they prove when teaching Fermat's Last Theorem, as it shows $m^4 + n^4 = r^4$ is impossible in nonzero integers.

Answer 2

As shown in the answer of Will Jagy, the problem reduces to showing that the equation $$ A^4+B^4=C^2, \tag{1} $$ does not possess a positive integer solution.

However, the fact that $(1)$ does not possess positive integer solutions can NOT be derived Fermat's Last Theorem. On the contrary, the fact that $(1)$ does not possess positive integer solutions implies Fermat's Last Theorem for $n=4$!

Here is why there is no solution to the Diophantine equation $(1)$.

The proof is based on a construction of an infinite descent, i.e., assuming that $C$ is the least positive integer for which the exists a triplet of positive integers satisfying $(1)$, we shall construct another such triplet with a small $C$.

Clearly, $(A^2,B^2,C)$ is a Pythagorean Triple (assuming without the loss of generality, $(A,B)=1$ and $B$ is even), so co-prime integers $a,b$ exist such that $$ A^2=a^2-b^2,\quad B^2=2ab\quad\text{and}\quad C=a^2+b^2. $$ The first equation becomes $A^2+b^2=a^2$, so $(A,b,a)$ is another Pythagorean Triple, with $b$ even. Thus co-prime integers $c,d$ exist such that $$ A=c^2-d^2, \quad b=2cd\quad\text{and}\quad\text{and}\quad a=c^2+d^2. $$ Substituting the expressions for $b$ to $B^2=2ab$, we have $$ B^2=4cd(c^2+d^2). $$ Now, since $c, d, c^2+d^2$ are clearly pairwise co-prime, they all are perfect squares. Hence, $$ c=e^2, \quad d=f^2\quad \text{and}\quad c^2+d^2=g^2. $$ Combining these expressions we obtain that $$ e^4+f^4=g^2, $$ with $g \le g^2 = a \le a^2 < C$.

Thus, assuming $C$ is minimally chosen, we found a smaller value for $C$.

Therefore, no solution.

Note. The most elementary proof of Fermat's Last Theorem for $n=3$ is also based on infinite descent. See here.

Answer 3

$ab=square\iff a=a_1^2c_1 and\space b=b_1^2c_2$ where $c_1 and \space c_2$ are square-free and $c_1c_2$ is a square.

This square product being assumed, if $a^2+b^2= square$ then $a_1^4c_1^2+b_1^4c_2^2= square$. Thus the equation $x^4+y^4=z^2$ could have a rational solution with $xyz\ne 0$. This is very known to be impossible (Fermat himself have proved it).