Find all $(x,y,z) \in \mathbb{N}$ such that $x^2 + y^2 =3(2023)^z +77$

Question

The question says Do not use any modular arithmetic.

My attempt:

First of all i think that these two things would help:

If $4 \mid x^2 + y^2 \implies 2\mid x$ And $y$. I’ve proved this one

If $7 \mid x^2 + y^2 \implies 7 \mid x$ And y. I don’t know how to prove this.

And it turns out that the only possible values to $(x,y,z)=(4,8,0)$.

By the way, I think it's a bit strange restriction (especially if you are allowed using divisibility but not modular arithmetic!). — richrow, Feb 08 '21 at 21:07
When i talk about positive integers i use $ \mathbb{N}^*$ @AsherAbraham — PNT, Feb 08 '21 at 22:20
@Asher2211: there are different conventions about whether $0$ is a natural number — J. W. Tanner, Dec 25 '24 at 22:45

score 2 · Accepted Answer · edited Dec 25 '24 at 22:47

2

We can write $x=7k+r$ where $r\in\{0,1,2,3,4,5,6\}$ so $$x^2 = 7(7k^2+2kr)+r^2$$ so $$x^2 = 7k'+r^2$$ Similary we have $$y^2 = 7l'+r'^2$$

Since $7\mid x^2+y^2$ we have now $$7\mid r^2+ r'^2$$

but this is possible only if $r=r'=0$ i.e. $7\mid x$ and $7\mid y$.

We see that if right side is divisible by 7 so $7\mid x$ and $7\mid y$ so $49$ divides RHS but that means that $49\mid 7$ if $z\geq 2$. So $z\leq 1$. Can you finish now?

edited Dec 25 '24 at 22:47

J. W. Tanner

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answered Feb 08 '21 at 20:49

nonuser

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I didn’t understand you ‘Spoiler’. How can $49 \mid 7$ ? – PNT Feb 09 '21 at 14:59
That can not be so z=<1 – nonuser Feb 09 '21 at 15:10
What is the relation between : $z\geq 2$ And $49 \mid 7$ – PNT Feb 09 '21 at 15:32
If $z\geq 2$ then $49\mid 3(2023)^z$, else that is not true. – nonuser Feb 09 '21 at 15:34
But in the case $z=1$ There is no solution . Because $4 \mid x^2 + y^2$ But $4\nmid 6146$ . – PNT Feb 09 '21 at 16:00
Ok.............. – nonuser Feb 09 '21 at 16:24

Find all $(x,y,z) \in \mathbb{N}$ such that $x^2 + y^2 =3(2023)^z +77$

1 Answers1