In 1752, Goldbach submitted the following conjecture to Euler?

[Math100]

Homework Statement

In 1752, Goldbach submitted the following conjecture to Euler: Every odd integer can be written in the form $p+2a^2$, where $p$ is either a prime or $1$ and $a\geq 0$. Show that the integer $5777$ refutes this conjecture.

Relevant Equations

None.

Proof:

Suppose $5777=p+2a^2$, where $p$ is either a prime or $1$ and $a\geq 0$.
Now we consider two cases.
Case #1: Suppose $p$ is a prime and $a\geq 0$.
Let $p=2$.
Then $5775=2a^2$.
Thus $a=\pm \sqrt{2887.5}$,
which contradicts the fact that $a\geq 0$.
Case #2: Suppose $p=1$ and $a\geq 0$.
Then $5776=2a^2$.
Thus $a=\pm \sqrt{2888}$,
which contradicts the fact that $a\geq 0$.
Therefore, the integer $5777$ refutes this conjecture.

 

[Office_Shredder]

What about all the other possible primes?

if $p=2$ then $p+2a^2$ is even, so the conjecture has to be referring to odd primes

 

[Math100]

So if $p$ is an odd prime, then $p=2k+1$ for some $k\in\mathbb{N}$?

 

[Math100]

And $5777=p+2a^2=2k+1+2a^2=2(k+a^2)+1$?

 

[fresh_42]

Homework Statement:: In 1752, Goldbach submitted the following conjecture to Euler: Every odd integer can be written in the form $p+2a^2$, where $p$ is either a prime or $1$ and $a\geq 0$. Show that the integer $5777$ refutes this conjecture.
Relevant Equations:: None.

Proof:

Suppose $5777=p+2a^2$, where $p$ is either a prime or $1$ and $a\geq 0$.
Now we consider two cases.
Case #1: Suppose $p$ is a prime and $a\geq 0$.
Let $p=2$.
Then $5775=2a^2$.
Thus $a=\pm \sqrt{2887.5}$,
which contradicts the fact that $a\geq 0$.

No. $p=2$ implies $2\,|\,(5777-2)$ which is the contradiction. $a$ has to be an integer. The sign is irrelevant since in case $a<0$ we automatically have $(-a)>0$ which would solve the equation.

Case #2: Suppose $p=1$ and $a\geq 0$.
Then $5776=2a^2$.
Thus $a=\pm \sqrt{2888}$,
which contradicts the fact that $a\geq 0$.
Therefore, the integer $5777$ refutes this conjecture.

Same as before. The contradiction is, that $a^2=2888$ implies $a\not \in \mathbb{Z}.$ The sign doesn't matter.

Now, you have ruled out $p\in \{1,2\}.$

However, we have to rule out all primes up to $5777.$ This is either troublesome, or we have to find a good idea of how to approach it. I would probably write a program, test all odd numbers for $p$, and check that those which solve $\sqrt{(5777-p)/2}\in \mathbb{Z}$ are not primes.

$A(n) =$Every odd integer $n$ can be written as $n=p+2a^2.$

E.g. $A(15)$ is solved by $(p,a)=(7,2)$ and $A(21)$ by $(p,a)=(3,3).$ In order to show that $A(5777)$ cannot be solved, we have to test all possible solutions $(p,a)$ and show that there is none. If I counted correctly, then there are $756$ possible primes $p$ which still have to be rejected.

Or test for $a=1,2,\ldots,53$ that $5777-2a^2$ isn't prime.

 

[Math100]

No. $p=2$ implies $2\,|\,(5777-2)$ which is the contradiction. $a$ has to be an integer. The sign is irrelevant since in case $a<0$ we automatically have $(-a)>0$ which would solve the equation.


Same as before. The contradiction is, that $a^2=2888$ implies $a\not \in \mathbb{Z}.$ The sign doesn't matter.

Now, you have ruled out $p\in \{1,2\}.$

However, we have to rule out all primes up to $5777.$ This is either troublesome, or we have to find a good idea of how to approach it. I would probably write a program, test all odd numbers for $p$, and check that those which solve $\sqrt{(5777-p)/2}\in \mathbb{Z}$ are not primes.

$A(n) =$Every odd integer $n$ can be written as $n=p+2a^2.$

E.g. $A(15)$ is solved by $(p,a)=(7,2)$ and $A(21)$ by $(p,a)=(3,3).$ In order to show that $A(5777)$ cannot be solved, we have to test all possible solutions $(p,a)$ and show that there is none. If I counted correctly, then there are $756$ possible primes $p$ which still have to be rejected.

Or test for $a=1,2,\ldots,53$ that $5777-2a^2$ isn't prime.

Is there another way to do this problem without writing a program? Because I have no knowledge/skill in writing a program. I am a total beginner in this.

 

[fresh_42]

Is there another way to do this problem without writing a program? Because I have no knowledge/skill in writing a program. I am a total beginner in this.

Well, it seems the shortest way is to consider $5777 -2a^2=p.$ That is, we have to check whether $p_0=5777-0\, , \,p_1=5777-2\, , \,p_2=5777-8\, , \,p_3=5777-18\, , \,p_4=5777-32\, , \,\ldots\, , \,p_{53}=5777-2\cdot 53^2$ contains a prime number. In order to conclude, that $5777$ is a counterexample, all these numbers $p_0\, , \,\ldots\, , \,p_{53}$ must be composite.

The second answer of
https://math.stackexchange.com/questions/1405982/a-false-conjecture-by-goldbach/3779005#3779005
is the best one. Consider the remainders of $a$ by division by $3$.
\begin{align*}
5777-2a^2=3\cdot 1925 +2 -2\cdot a^2 =\begin{cases}0&\text{ if } a=3\cdot q+1 \text{ or }3\cdot q +2\text{ for some }q\\ 2&\text{ if } 3\,|\,a\end{cases}
\end{align*}
In the first case, we have $3\,|\,(5777-2a^2)$ which is only prime for $5777-2a^2=3.$ But there is no such integer $a.$ In the second case, we have $a=3q$ for some $q\in \mathbb{N}$ and
$$
5777-2a^2=320\cdot 18 + 17 - 2\cdot 3^2\cdot q^2= 18\cdot (320 - q^2)+ 17
$$
Hence, if $5777-2a^2=p$ is a prime number, then $p$ has to leave the remainder $17$ by a division by $18.$ So we have to check all prime numbers among $17\, , \,35\, , \,53\, , \,\ldots\, , \,5777.$ These shouldn't be many which are of the form $p=18\cdot q+17$ and prime. You can compare these numbers with the list here:
https://de.wikibooks.org/wiki/Primzahlen:_Tabelle_der_Primzahlen_(2_-_100.000)

I still think an Excel sheet would be the fastest way to check the $54$ numbers for $a$. And if you do not have Excel, then you can use Open Office or Libre Office.

 

[fresh_42]

Done with OpenOfficeCalc:

a​	2a^2​	p=5777-2a^2 prime?​
0​	0​	5777​
1​	2​	5775​
2​	8​	5769​
3​	18​	5759​
4​	32​	5745​
5​	50​	5727​
6​	72​	5705​
7​	98​	5679​
8​	128​	5649​
9​	162​	5615​
10​	200​	5577​
11​	242​	5535​
12​	288​	5489​
13​	338​	5439​
14​	392​	5385​
15​	450​	5327​
16​	512​	5265​
17​	578​	5199​
18​	648​	5129​
19​	722​	5055​
20​	800​	4977​
21​	882​	4895​
22​	968​	4809​
23​	1058​	4719​
24​	1152​	4625​
25​	1250​	4527​
26​	1352​	4425​
27​	1458​	4319​
28​	1568​	4209​
29​	1682​	4095​
30​	1800​	3977​
31​	1922​	3855​
32​	2048​	3729​
33​	2178​	3599​
34​	2312​	3465​
35​	2450​	3327​
36​	2592​	3185​
37​	2738​	3039​
38​	2888​	2889​
39​	3042​	2735​
40​	3200​	2577​
41​	3362​	2415​
42​	3528​	2249​
43​	3698​	2079​
44​	3872​	1905​
45​	4050​	1727​
46​	4232​	1545​
47​	4418​	1359​
48​	4608​	1169​
49​	4802​	975​
50​	5000​	777​
51​	5202​	575​
52​	5408​	369​
53​	5618​	159​

Now check the last row for primes. Strikeout all which ends on 0 or 5, then all with digit sum divisible by 3 and finally the few that are left.

 

[Math100]

But I want to know, why choose $3$ by division in this problem? With two cases of $a\mid 3$ and $a\nmid 3$?

 

[fresh_42]

But I want to know, why choose $3$ by division in this problem? With two cases of $a\mid 3$ and $a\nmid 3$?

This is just the smallest prime (2 doesn't get us very far) with only a few possible remainders. You can also consider any other prime to narrow the number of cases that have to be checked. But I think, my approach with the 54 possible a's is shorter. After I eliminated the ones dividable by 3 or 5 in my list above, I was left with only 11 numbers to check.

I had only 4 cases with prime factors higher than 13. The factors 7,11, and 13 can be done with a calculator, the remaining four with WolframAlpha.

5777-2a^2​	divisor​	​	​
5777​	​	53​	109​
5775​	5​	​	​
5769​	3​	​	​
5759​	13​	​	​
5745​	5​	​	​
5727​	3​	​	​
5705​	5​	​	​
5679​	3​	​	​
5649​	3​	​	​
5615​	5​	​	​
5577​	3​	​	​
5535​	5​	​	​
5489​	11​	​	​
5439​	3​	​	​
5385​	5​	​	​
5327​	7​	​	​
5265​	5​	​	​
5199​	3​	​	​
5129​	​	23​	223​
5055​	5​	​	​
4977​	3​	​	​
4895​	5​	​	​
4809​	3​	​	​
4719​	3​	​	​
4625​	5​	​	​
4527​	3​	​	​
4425​	5​	​	​
4319​	7​	​	​
4209​	3​	​	​
4095​	5​	​	​
3977​	​	41​	97​
3855​	5​	​	​
3729​	3​	​	​
3599​	​	59​	61​
3465​	5​	​	​
3327​	3​	​	​
3185​	5​	​	​
3039​	3​	​	​
2889​	3​	​	​
2735​	5​	​	​
2577​	3​	​	​
2415​	5​	​	​
2249​	13​	​	​
2079​	3​	​	​
1905​	5​	​	​
1727​	11​	​	​
1545​	5​	​	​
1359​	3​	​	​
1169​	7​	​	​
975​	5​	​	​
777​	3​	​	​
575​	5​	​	​
369​	3​	​	​
159​	3​	​	​

 

[Math100]

This problem is a bit tedious.

 

[TeethWhitener]

There are some cheeky ways to eliminate a few of the $a$’s. For instance, a square number always ends in 0, 1, 9, 6, or 5, and doubling a square will give numbers ending in 0, 2, and 8. Notice that 5777 minus a number that ends in 2 will give a number that ends in 5. This method will allow you to eliminate (from my count) 21 of the 53 numbers (I’ll leave you to figure out which ones).

Another trick is to notice that 5777 is $2 \mod 3$. So subtracting another number which is $2\mod3$ will give a number which is divisible by 3. If you know a bit of modular arithmetic, you can quickly figure out the conditions on $a$ that make $2a^2\equiv2\mod3$. This should eliminate most of the rest of the possible $a$’s. By my count, you should have about 10 possible $a$’s to check by hand when the dust settles.

 

[nuuskur]

Please don't use such loaded (and even clickbaity) titles. Be concise and state the problem.

If $a\equiv 0 \pmod{3}$, then $p\equiv 17\pmod{18}$ cuts down the cases to be checked significantly. $a\equiv 1,2\pmod{3}$ leads to $3\mid p$, but $p\neq 3$.