10
$\begingroup$

"This proof was found by Sergey Markelov when yet in high school. In the decimal system, a square of an integer may only end in one of the following digits: $0$, $1$, $4$, $5$, $6$, $9$ whereas twice a square may only end with $0$, $2$, $8$. Thus assuming $a^2=2b^2$, both $a$ and $b$ may only end with $0$. This triggers an infinite descent which proves that this is also impossible."

What exactly is the infinite descent here? I can see that the only solution is $a=0$, $b=0$, but I don't see how that shows that there exist smaller $a$, $b$.

$\endgroup$
3
  • 1
    $\begingroup$ Hint: If they both end with a $0$, they both are divisible by $10$. $\endgroup$
    – Gary
    Apr 22, 2020 at 6:37
  • 2
    $\begingroup$ @jamie thank you for sharing this beautiful proof. :) $\endgroup$ Apr 22, 2020 at 7:03
  • $\begingroup$ Just for the record, its proof no 14 here cut-the-knot.org/proofs/sq_root.shtml $\endgroup$
    – user600016
    Apr 22, 2020 at 7:15

2 Answers 2

3
$\begingroup$

I was struggling to see how the last digit of $b^2$ must be 0. I believe the answer is that it is not necessary, it could be 0 or 5. It can be seen that the last digits of $a^2,b^2$ must be in 0,1,4,9,6,5 , so cannot be 2 or 8,

$$ a^2\equiv 0 \mod 10$$ $$ 2b^2\equiv 0 \mod 10$$ Since $2|10$, $$ b^2\equiv 0 \mod 5$$ Then since 5 is prime, we can make the argument that the factor of 5 did not come from squaring. Therefore if true for $b,a$ we can divide by 5 to find it is also true for two smaller integers, leading to a contradiction by infinite descent due to the well ordering principle.

$\endgroup$
2
$\begingroup$

Assume for contradiction that there are non zero natural numbers $a, b$ such that $a^2 = 2 b^2$. Consider $a \neq 0 \in \mathbb{N}$ minimal such that there is $b \in \mathbb{N}$ (wlog) such that $a^2 = 2 b^2$. As observed, the last digit of $a^2$ and $2b^2$ are $0$, so that both are divisible by $5$.

Hence, both $a$ and $b$ are divisible by $\textbf{5}$ (and not necessarily by $10$, as noted by @jamie in his excellent answer).

Then $a' := \frac{a}{5}$ and $b' := \frac{b}{5}$ also satisfy $a'^2 = 2 b'^2$. It is a contradiction since $a' < a$.

Btw I skiped the "infinite descent" but it amounts to this : if any $a, b$ satisfy $a^2 = 2 b^2$ then $a' := \frac{a}{5}$ and $b' := \frac{b}{5}$ also satisfy $a'^2 = 2 b'^2$ and eventualy you can't divide by $5$ anymore.

$\endgroup$
2
  • 1
    $\begingroup$ Please see my answer, I think you(and the quoted proof) were wrong to assume that the last digit of $2b^2$ is 0 $\endgroup$
    – jamie
    Apr 22, 2020 at 16:01
  • $\begingroup$ @jamie Well done, I included your remark. $\endgroup$ Apr 23, 2020 at 2:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.