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For the following question: Given $ \equiv 4 \pmod{13}$, and $ \equiv 9\pmod{13}$. If $\equiv ^2+^2\pmod{13}$ with $0\leq c \leq 12$ . Find the value of $c$.

Here is my attempt :

$a^2 \equiv 4 \cdot 4 \pmod{13}$

$a^2 \equiv 16 \pmod{13}$

$b^2 \equiv 9\cdot 9 \pmod{13}$

$b^2 \equiv 81 \pmod{13}$

then

$a^2+b^2 \equiv 81+16 \pmod{13}$

$a^2+b^2 \equiv 97 \pmod{13}$

then

$c \equiv a^2+b^2 \equiv 97 \equiv 6 \pmod{13}$

and hence $c=6$. Is my solution correct? More specifically, for $a^2$ and $b^2$ where I have multiplied the congruent relations of $a$ and $b$ by themselves, is that right?

Arturo Magidin
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  • What are $a2$ and $b2$? – Bernard Nov 15 '21 at 20:42
  • do you mean $c \equiv 2a + 2b \mod 13$ or $c\equiv a^2 + b^2 \mod 13$? I suppose the latter. – Student Nov 15 '21 at 20:42
  • @Bernard . edited . Sorry – AAA Nov 15 '21 at 20:44
  • Please use proper MathJax. – Arturo Magidin Nov 15 '21 at 20:44
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    You've been a user here for three and a half years. Don't you think it's time you learnt MathJax? – Shaun Nov 15 '21 at 20:44
  • @Student yes. This was a typo – AAA Nov 15 '21 at 20:44
  • Also, your first line is $\mod 13$, the second becomes $\mod 3$. I suppose that's a typo too. Pleas follow Shaun's suggestion and use MathJax to make questions more readible. – Student Nov 15 '21 at 20:46
  • Yes, it's correct (mod typos), by using the Congruence Laws in the linked dupe. Easier: $,a\equiv 4\Rightarrow a^2\equiv 4^2\equiv 3,,$ $, b\equiv 9\equiv -4\Rightarrow b^2\equiv (-4)^2\equiv 3,,$ so $,a^2+b^2\equiv 3+3\ \ $ – Bill Dubuque Nov 15 '21 at 20:50
  • Also, not related to this question, but after looking at your profile I couldn't help but notice that you don't accept answers. If answers are helpful, you should accept them :) – Student Nov 15 '21 at 20:52
  • You can find many more examples of congruence arithmetic in the Linked posts in the dupe, and well as searching in the modular arithmetic tag. – Bill Dubuque Nov 15 '21 at 20:52
  • More generally $,a+b\equiv 0\Rightarrow b\equiv -a\Rightarrow a^2+b^2\equiv a^2+(-a)^2\equiv 2a^2\ \ $ – Bill Dubuque Nov 15 '21 at 20:55
  • @Student: If you don't know how to properly type it up in MathJax, then kindly do not try to "help". That is a true mess. – Arturo Magidin Nov 15 '21 at 20:57
  • This is probably an opportune time to mention meta-cheating. Since you are not given the exact values of $a$ or $b$, then you can presume that the answer is independent of the exact values. This means (to a certain extent) you can ignore Number Theory and assume that $a = 4, b = 9, ~$ exactly. That is, by meta-cheating, if you compute with those exact values for $a,b$, and if the problem is solvable, then your computation must achieve the (unique) intended answer. I mention this because in both high school and college math, meta-cheating is a powerful weapon. – user2661923 Nov 15 '21 at 22:53
  • @ArturoMagidin I kept the original structure of OP's answer, just corrected typos, added missing exponentsigns and added '$'. Looking at your edits, I learned about the pmod-command, so thank you! – Student Nov 16 '21 at 06:35

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