Absurd Prime

Let the positive integers be $a$ and $b$ , then we've, $a^{2}-b^{2}=p$ , or $(a+b)(a-b)=p$ . Now, since $p$ is a prime, thus, it can't have two prime factors other than $1$ or $p$ . Thus, one of the two factors, must be equal to $1$ . Now, since, $(a-b)$ is smaller of the two, thus, it must be equal to $1$ . Thus, $(a-b)=1$ and $(a+b)=p$ . Solving the former, we get, $a=b+1$ . Substituting in second equation, $b=(p-1)/2$ and $a=(p+1)/2$ . Thus $ab=(p^{2}-1)/4$ . Now, since $ab$ is an integer, therefore, $p^{2}-1$ is div. by $4$ . or, $4$ divides $(p^{2}-1)+136=p^{2}+135$ . [Since, $136$ is divisible by $4$ .]. Thus, on dividing $p^{2}+138$ by $4$ , the remainder is $3$ . [since, $p^{2}+135$ is div. by $4$ ].

It is clear that $2$ cannot be expressed as the difference of two squares of positive integers (as it is congruent to $2$ modulo $4$ ), so $p$ must be odd. Hence, $p^2 \equiv 1 \pmod{4}$ . Therefore, $p^2+138 \equiv 139 \pmod{4}$ , and $139 \equiv 3 \pmod{4}$ , so it follows that $p^2+138 \equiv 3 \pmod{4}$ .

$(p=a^2-c^2)\Rightarrow (p\neq 2)\Rightarrow (p=2 k+1) \\ \therefore\ p^2+138=p^2+4\times 34+2 \\ =(2k+1)^2+4\times 34+2 \\ =4 k^2+4k+4\times 34+2+1 \\ =4(k^2+k+34)+3 \\ \therefore\ \boxed{(p^2+138) \mod 4 = 3}$

Any prime no. when divided by 4 gives remainder 1. Hence, p^2/4 gives rem 1 and 138/4 gives rem 2. Therefore , overall rem=rem of (1+2)/4=3.

It also happens to be that you have 3 guesses and there are 3 possible choices for a remainder of 4, 1,2, and 3. So you can theoretically guess all possible answers and get it right. However, I did solve it the same way as souhardya. Just saying that this question is flawed because of this reason.

We know that $p = m^2 - j^2 = (m-j)(m+j)$ where $m, j$ are positive integers.

Since $p$ is prime, we have:

$\begin{cases} m-j=1 \\ m+j=p \end{cases}$ $\Leftrightarrow 2m-1=p$

Squaring both sides, we get:

$p^2 = 4m^2-4m+1 \Rightarrow p^2 = 4(m^2 - m) + 1$

which means that the expression above leaves a remainder of 1 when divided by 4, in other words:

$p^2 \equiv 1 \quad(mod \quad4)\quad \quad \quad [*]$

We know that

$138 \equiv 2 \quad (mod \quad 4)\quad \quad \quad[**]$

adding $[**]$ and $[*]$ , gives us:

$p^2 + 138 \equiv 3 \quad (mod \quad 4)$

Therefore, the remainder of the expression when divided by 4 is $\boxed{3}$

Find the easiest example of a prime number that fits the description. 3 = 2^2 - 1^2. 3^2 +138 = 147. Remainder of 147 / 4 = Remainder of 47/4 = 11 Remainder 3.

You get 3 guesses on the answer of mod(x,4), which can only have 3 possible solutions regardless of all the hints in the question. So you shouldn't possibly get this wrong.

The two integers are 9 & 8 (9×9) - (8×8) =17 17×17= 289 289+138 =427 427÷4 gives remainder 3

The Prime Number can't be 2 as if 2=x^2-y^2, then X and Y wouldn't be integers, Therefore it is an ODD PRIME NUMBER in the form of 4k+1 or 4k-1... hence (4k-or+1)^2 + 4*34+2 = 4(p+34)+3.....Therfore Reminder is 3

check for 4^2 - 3^2 = 7 , now 138 + 7^2 = 187 , reminder (187/4) = 3

p = 5 = 3^2 - 2^2 138+25 = 163 163/4 = 40 and remainder = 3

minimum difference of 2 squares=3. so, p=1(mod 2). (2k+1)^2=1(mod 4) . p should be the difference of a odd and a even square. (2k)^2=0(mod 4)=>p=1(mod 4)=>p^2=1(mod 4) 138=2(mod 4) p^2+138=1+2=3(mod 4) remainder=3

Since two prime numbers are such that,square of first should be bigger than the second one,also after substraction when added to 138 and divided by 4 shold come remainder as whole number. considering X & Y I found 7 and 6 are such prime numbers and when solved , my answer is 3 remainder. K.K.GARG,India

Let a^2 - b^2 = p (a+b)(a-b) = p

Difference of the two numbers is given by 2b, which is twice the number, b.

Now the only numbers that satisfy these conditions are a = 3 and b = 2, because 3^2 - 2^2 = 5, a prime which is good.

Next (3+2)-(3-2) = 4, which is 2(2), 2b, again, great.

Knowing that p = 5, the solution follows.

the picture is epic

If $p$ is a prime number, then $p$ must be either 2 or any other odd number . since the difference the squares of two positive integers can never be 2 , then p must be an odd number.

If $p$ is an odd number, the number can be written as $2x+1$ for a certain value of $x$ . In other words, we can rewrite our equation as: $(2x+1)^2 + 138 \equiv N \pmod{4}$ $or$ , $4x^2 + 4x + 1 + 138 \equiv N \pmod{4}$ $or$ , $0 + 0 + 1 + 2 \equiv N \pmod{4}$ $or$ , $3 \equiv N \pmod{4}$ .

Therefore, our answer is $\boxed{3}$ .

If p = 2 then 138 + 4 = 142 and 142 / 4 = 35 + 1/2 so remainder is 2/4 If p is greater than 2 then the remainder is 3/4

I tested 2 and 1 ants I got 3 A’s the remainder

NB: This isn't a formal proof, and relies on the phrasing of the question.

$p^2 = x^2 - y^2$ with all three numbers being integers. Rearranging gives $p^2 + y^2 = x^2$ , which we can use to conceptualize the three numbers as side lengths of a right triangle (x being the hypotenuse). Since all three numbers are integers, we have a pythagorean triple.

This is where the fudgery comes in. Since we know there exists more than one pythagorean triple, and since there is no way know which, if any, the author had in mind when writing the question, this implies to me that it doesn't matter which one is used going forward (i.e. all triples will give the same answer to the question).

Pick the easy triple: 3, 4, 5. The hypotenuse must be 5, and therefore $3 = p$ , as 4 is not prime. The answer then follows from a simple calculation.

let us take p=(3-2)(3+2) i.e. smallest possibility for difference of square of 2 positive no.s yielding a prime no. then => p=5 & p^2+138=25+138=163 thus remainder is 3

This isn't really a "solution", but I just used Pythagoras' theorem, a2 + b2 + c2, or simply, 3 squared plus 4 squared equals 5 squared. Assuming the prime p had to be either a or b, and it can't be 4, then it must be 3. 3 squared plus 138 equals 147, divided by 4, gives the remainder 3.

$if\quad we\quad consider\quad a\quad prime\quad "p"\quad then\quad the\\ \quad following\quad cases\quad can\quad occur\\ case:1\quad \\ p\quad \equiv \quad 1\quad (mod\quad 4)\\ { p }^{ 2 }\quad \equiv \quad 1\quad (mod\quad 4)\\ case\quad :\quad 2\\ p\quad \equiv \quad 3\quad (mod\quad 4)\\ { p }^{ 2\quad }\equiv \quad 9\quad (mod\quad 4)\quad and\\ \quad 9\quad \equiv \quad 1\quad (mod\quad 4)\quad \\ { p }^{ 2 }\quad \equiv \quad 1\quad (mod\quad 4)\\ 138\quad \equiv \quad 2\quad (mod\quad 4)\quad \\ { p }^{ 2 }\quad +\quad 138\quad \equiv \quad 3\quad (mod\quad 4)\\ hence\quad remainder\quad 3$

Claim1: $p \neq 2$

Call the two positive integers $a$ and $b$ , then $p = a^2 - b^2 = (a + b) (a - b)$ .

Since p is prime, then one of the two terms is $p$ , and the other is 1, since $a-b$ is the smaller term, $a - b = 1$ and $a+b = p$ .

If $p = 2$ then $a = b = 1$ , but $a - b = 1 - 1 = 0 \neq 1$ , contradiction. Therefore $p \neq 2$ . QED

Claim 2: $p \% 4 = 1$ or $p \% 4 = 3$

Since $p$ is prime and $p \neq 2$ , then $p \% 4 \neq 0$ and $p \% 2 \neq 0$ , hence $p \% 4 \neq 2$ .

This means that $p \% 4 = 1$ OR $p \% 4 = 3$ QED

Main Proof

Let $r = p \% 4$ , so $r= 1$ or $r = 3$

Let $p = k + r$ . Note that $k \% 4 = 0$ .

\(p^2 = (k + r)^2 = k^2 + 2kr + r^2

Since \(k \% 4 = 0\), $k^2 \% 4 = 0$ , and $2kr \% 4 = 0$ . This leaves us with $p^2 \% 4 = 0 + 0 + r^2$

If $r = 1$ then $r^2 \% 4 = 1 \% 4= 1$

If $r = 3$ then $r^2 \% 4 = 9 \% 4= 1$

Hence $r^2 \% 4 = 1$ always. Thus $p^2 \% 4 = 0 + 0 + 1 = 1$

Finally $138 \% 4 = 2$ , and so $(p^2 + 138) \% 4 = (1 + 2) \% 4 = 3$

p is a prime number so it should be in the form 4n+1 or 4n+3 (numbers in the form 4n&4n+2 are multiples of 2) so p^2 should be in the form 4n+1 so the reminder is 1+2(2 is the reminder when 138 is divided by 4)

its 3 because everythingis