Question on divisibility by 3. Prove that a²-36=8 is not true for any natural numbers a and b.

Asked Feb 01 '22 at 11:04

Active Feb 02 '22 at 13:14

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Prove that there are no natural numbers a and b such that a²-36b=8.

A hint is given in the solutions that we should check the remainders by mod 3. But do we have to take remainders of whole expression or a and b seperately ?

edited Feb 02 '22 at 13:14

asked Feb 01 '22 at 11:04

Rishabh Shrivastava

4

What is the b? Where is that? – Mikasa Feb 01 '22 at 11:06
You mean $a^2 - 36 b = 8$? If it's so : 1) Prove that $a$ is even and put $a = 2 c$. 2) Prove that $c$ is even. 3) Deduce. – Essaidi Feb 01 '22 at 11:17
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@Essaidi I think that the OP wanted to write $a^2-3b=8$. – Mikasa Feb 01 '22 at 11:18
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In both cases , modulo $3$ the equation becomes $a^2=2$ which has no integer solution. – Peter Feb 01 '22 at 11:24
$a^2\equiv 36+8\equiv 2\pmod{3}$ is unsolvable since $2$ is not a square $!\bmod 3,$ - see the linked dupes. – Bill Dubuque Feb 01 '22 at 17:06
The original question is a²-36b=8. – Rishabh Shrivastava Feb 02 '22 at 13:15

Question on divisibility by 3. Prove that a²-36=8 is not true for any natural numbers a and b.

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