Mr 7 challange you!

The squares modulo 7 are $\begin{aligned} 0^2 & \equiv 0 \\ 1^2 \equiv (-1)^2 & \equiv 1 \\ 2^2 \equiv (-2)^2 & \equiv -3 \\ 3^2 \equiv (-3)^2 & \equiv 2 \end{aligned}.$ The only way in which we can add two of these squares to obtain $0$ modulo 7 is $0^2 + 0^2 \equiv 0 + 0 \equiv 0.$ In other words, both squares are squares of multiples of 7.

every square have as a remainder mod 7 only one of the values 0 or 1 or 2 or 4, 7 can be write only like the sum of the (7 and 0) or (6 and 1) or (5 and 2) or (4 and 3), so (7 and 0) is the only valid way to verified the condition of the exercice, a² and b² must be each devisible by 7 , like 7 is a prime number (or in genetral a number with an odd exponent on all prime factors) , a and b must be each devisible by 7 .

Since 7 is prime, this is actually equivalent to the statement $-1 \text{ is a quadratic nonresidue modulo } 7$

For moderately sized primes or primes of special form, it would be best to investigate this using the tools for quadratic residues, but in our case, we can just list the nonzero quadratic residues as $1^2 \equiv 1 \pmod{7} \\ 2^2 \equiv 4 \pmod{7} \\ 3^2 \equiv 2 \pmod{7}$

Note that $-1 \equiv 6 \pmod{7}$ doesn't appear, so the answer is $\boxed{\text{True}}$

Suppose a = 7m+r, 0<r<7 and b= 7n +s, 0<s<7. Then a^2 + b^2 = (7m+r)^2 + (7n+s)^2 = (49m^2 + 14mr + r^2) + (49n^2 + 14ns + s^2) = 49(m^2+n^2+ 14(mr +ns) + (r^2+s^2) which is divisible by 7 iff r^2+s^2 is divisible by 7. So now we do the same for r^2+s^2 and we obtain that this is divisible by 7 iff r'^2+s'^2 is divisible by 7. This will go on for ever, but it can't since with finite integers we can't lower them for ever, So the answer is yes, a and b must be divisible by 7.

$a^2+b^2 = 7n$ ; this is the given fact, $n$ is a positive integer.

One can assume that $a = 7r$ and $b = 7t$ ; where $r$ and $t$ are both positive integers; this says that $a$ & $b$ are both divisible by $7$ .

Applying the given fact: $a^2+b^2 = 7n$ $\rightarrow$ $(7r)^2+(7t)^2 = 7n$ $\rightarrow$ $7^2(r^2+t^2) = 7n$ $\Rightarrow$ $\boxed{7(r^2+t^2) = n}$

The last equation shows that yes both $a$ and $b$ can always be divisible by $7$ since for some integer values to each of $r$ & $t$ , the variable $n$ will be an integer as well (and as it should be).

From Euclid formula, pythagorean triples are generated from choices of m, n where m>n>0.

a=m^2-n^2

b=2mn

c=m^2+n^2

This formula includes all primitive triplets.

For c^2 to be divisible by 7, c itself must be divisible by 7 since 7 is prime.

Thus one must find any number c, that is divisible by 7 that is the sum of two squares m^2+n^2.

Consider modulo 7 arithmetic. There are 7 digits, 0,1,2,3,4,5,6

The squares of these digits modulo 7 are 0,1,4,2,2,4,1 respectively.

There is no way to combine any two these numbers with addition modulo 7 to obtain 0.

Therefore no two squares sum to any number divisible by 7. Therefore no pythagorean triple may be generated by Euclid formula such that c^2 is divisible by 7. Since Euclid formula includes all primitive triples, there are no primitive triples with c^2 divisible by 7. Any non-primitive triple with c^2 divisible by 7 must be a scaled version of a primitive triple where the scaling factor is divisible by 7, but in that case, a and b are also scaled by the same factor and must also be divisible by 7.

Example: 21^2+28^2=35^2 is really primitive 3,4,5 scaled by 7.

When we divide a square number by 7, the possible remainders are 0,1,2,4. So, when we divide a^2 by 7 and b^2 by 7 individually ,we get the remainders 0,1,2 or 4. Possible remainders: when a^2 is divided by 7 are [0,1,2,4] and when b^2 is divided by 7 are [0,1,2,4]. Given that a^2+b^2 is divisible by 7,means the remainder is 0. So if we take all possible remainders and their sum pair wise ,should be divisible by 7. Out of all possibilities only 0+0=0 is divisible by 7. Therefore a and b always divisible by 7.

If $a^{2}$ + $b^{2}$ is divisible by $7$ , we can let

$a^{2}$ = $7k$

$b^{2}$ = $7j$

for some positive integers $k$ and $j$ .

Well actually, for $a$ and $b$ to be positive integers, $k$ and $j$ must be in the form $7m^{2}$ , where $m^{2}$ is a perfect square, so that

$a$ = $\sqrt{7k}$ = $\sqrt{7(7m^{2})}$ = $\sqrt{(7m)^{2}}$ = $7m$

$b$ = $\sqrt{7j}$ = $\sqrt{7(7m^{2})}$ = $\sqrt{(7m)^{2}}$ = $7m$ .

In this case, $a$ and $b$ are always divisible by $7$ . Apparently, it would still hold true if we replace $7$ with any positive integer $n$ .

Any prime divisor of a^2 + b^2 is of the form 4k + 1. Ed Gray

if u consider a circle with a radius that's an integer multiple of 7 then any angle on that circle constructed from two lines through the center with integer solutions for its sin and cos will also be divisible by 7 because everything is just the unit circle scaled by some multiple of 7

Gauss theorem is necessary to go quickly and efficiently though the squares(^²) of this problem : if "a" divides b $\times$ c and "a" coprime with b then a divides c. In our case, if 7 divides a², then (first case, reduction ad absurdum, if 7 is coprime with "a" then 7 is coprime with a². Absurd. So only second case : 7 has a common divisor with a, that is 7.)

The solution needs to list the a² mod 7 : if a €{1,2,3,4,5,6} (or congruent to them ; anyway, not 0, not multiple of 7)≡{1,2,3,-3,-2,-1}, then a²€{1,4,9,9,4,1}, congruent [mod 7] to {1,4,2,2,4,1}. Also b² congruent to €{1,4,2} if b not multiple of 7. The addition of a²+b² (really tried every sum of last two equal ensembles) is congruent to €{2,5,3,1,6,4} and gives not 0. Contradiction with a²+b² congru 0 [mod 7] (reductio ad absurdum).

Or by contraposition "a not multiple of 7, nor b" => "a²+b² not≡0 [mod 7]". So "a²+b²≡0[7]" => a² or b² multiple of 7. 7 is prime number, so a or b multiple of 7. Then a² or b² also. If a² multiple of 7, then b²=7n-a² also ; if b² multiple of 7, then a²=7n-b² also. * a²+b² is a multiple of 7. *

$\text{Solution in Algebra}$

$\begin{aligned} a^{2}+b^{2} &= 7k \\ a &= \sqrt{7k-b^{2}} \\ \end{aligned}$ Assume that: $\begin{aligned} a+b &= c \\ \sqrt{7k-b^{2}} &= c-b \\ 7k-b^{2} &= c^{2}-2bc+b^{2} \\ c^{2} &= -2b^{2}+2bc+7k \\ c^{2} &= 2(a^{2}-7k)+2bc+7k \\ 0 &= c^{2}-2bc-2a^{2}+7k \\ c &= \frac{2b+\sqrt{(-2b)^{2}-4(-2a^{2}+7k)}}{2} \\ c &= b+\sqrt{b^{2}+2a^{2}-7k} \\ c &= b+\sqrt{b^{2}+2a^{2}-(a^{2}+b^{2})} \\ c &= b+\sqrt{a^{2}} \\ c &= a+b \\ \end{aligned}$

$\large \text{Thug Life LOL}$

$a^2+b^2=7n$ so $(a+bi)(a-bi) =7n$ so $n=\frac{(a+bi)(a-bi)}{7}$ and because $n$ is an integer, $\frac{(a+bi)(a-bi)}{7}$ must be an integer so $7$ must divide it. Either $a+bi$ or $a-bi$ contain $7$ or a multiple of $7$ or else the product won't give a $7$ because it's prime, so either $\frac{(a+bi)}{7}$ or $\frac{(a-bi)}{7}$ gives an integer. So either $(\frac{a}{7}+\frac{bi}{7})$ is an integer or $(\frac{a}{7}-\frac{bi}{7})$ is an integer so $a$ must be divided evenly by $7$ and so is $b$ because neither $i$ nor $-1$ are divisible by $7$

You can also get the solution with the Gaussian numbers (x + y i, x,y integers). Since numbers of the form 3 + 4k (where k is an integer) are prime elements, 7 is a prime element. Since a^2 + b^2 = (a + b i)(a - b i) and a^2 + b^2 = 7 x. 7 must be part of (a + b i) or (a - b i), thus it has to be part of a and b, since there is a unique factorization in the Gaussian numbers.

every square have as a remainder mod 7 only one of the values 0 or 1 or 2 or 4, 7 can be write only like the sum of the (7 and 0) or (6 and 1) or (5 and 2) or (4 and 3), so (7 and 0) is the only valid way to verified the condition of the exercice, a² and b² must be each devisible by 7 , like 7 is a prime number (or in genetral a number with an odd exponent on all prime factors) , a and b must be each devisible by 7 .

I used to use this property in problems of PRMO, RMO, NMTC directly. 😂😂

Simply look in Mod 7............