Let’s suppose $y$ is such that $144y^2 + 3600$ is divisible by 25 and results in a square.

["Let $ y $ Be Such That $ 144y^2 + 3600 $ Is Divisible by 25 and Results in a Perfect Square", "In advanced algebra and number theory, there are intriguing problems involving quadratic expressions and their divisibility properties. One such compelling case is determining values of $ y $ for which the expression $ 144y^2 + 3600 $ is both divisible by 25 and yields a perfect square. This exploration reveals deep insights into modular arithmetic, squares modulo powers, and Diophantine equations.", "---", "### Understanding the Problem", "We are given the expression:", "$$\n144y^2 + 3600\n$$", "We seek integer values of $ y $ such that:", "1. $ 144y^2 + 3600 \equiv 0 \pmod{25} $\n2. $ 144y^2 + 3600 = k^2 $ for some integer $ k $", "In other words, the expression must simultaneously be divisible by 25 and a perfect square.", "---", "### Step 1: Simplify the Divisibility Condition", "We start with divisibility by 25:", "$$\n144y^2 + 3600 \equiv 0 \pmod{25}\n$$", "Reduce coefficients modulo 25:", "- $ 144 \mod 25 = 144 - 5 \cdot 25 = 144 - 125 = 19 $\n- $ 3600 \mod 25 = 0 $ (since $ 25 \ imes 144 = 3600 $)", "Thus, modulo 25, the expression becomes:", "$$\n19y^2 \equiv 0 \pmod{25}\n$$", "Since 19 is coprime to 25 (i.e., $ \gcd(19,25) = 1 $), we can multiply both sides by the modular inverse of 19 modulo 25.", "Let’s find $ 19^{-1} \mod 25 $:", "We solve $ 19x \equiv 1 \pmod{25} $. Trying values or using the extended Euclidean algorithm:", "- $ 25 = 1 \cdot 19 + 6 $\n- $ 19 = 3 \cdot 6 + 1 $\n- Back-substitute: $ 1 = 19 - 3(25 - 19) = 4 \cdot 19 - 3 \cdot 25 $", "So $ 19^{-1} \equiv 4 \pmod{25} $", "Multiply both sides:", "$$\ny^2 \equiv 0 \cdot 4 \equiv 0 \pmod{25}\n\Rightarrow y^2 \equiv 0 \pmod{25}\n$$", "Thus, $ y \equiv 0 \pmod{5} $. So $ y = 5k $ for some integer $ k $.", "---", "### Step 2: Substitute $ y = 5k $ into the Original Expression", "Now plug $ y = 5k $ into $ 144y^2 + 3600 $:", "$$\n144(5k)^2 + 3600 = 144 \cdot 25k^2 + 3600 = 3600k^2 + 3600 = 3600(k^2 + 1)\n$$", "We require this to be a perfect square:", "$$\n3600(k^2 + 1) = k_{\ ext{new}}^2\n$$", "Factor 3600:", "$$\n3600 = 60^2 = (2^3 \cdot 3 \cdot 5)^2 = 2^6 \cdot 3^2 \cdot 5^2\n$$", "So:", "$$\n3600(k^2 + 1) = (60)^2(k^2 + 1) = m^2 \quad \ ext{for some integer } m\n\Rightarrow k^2 + 1 = \left( \frac{m}{60} \right)^2\n\Rightarrow k^2 + 1 = \left( \frac{a}{d} \right)^2 \ ext{ for some integer } a,d\n$$", "But instead, observe:", "For $ 3600(k^2 + 1) $ to be a perfect square, since 3600 is already a perfect square, $ k^2 + 1 $ must be a perfect square multiplied by a square factor — more precisely, $ k^2 + 1 $ must be a perfect square times a unit, but since $ k^2 + 1 \in \mathbb{Z}^+ $, we can write:", "Let:", "$$\nk^2 + 1 = s^2 \cdot t^2 \quad \ ext{but simpler: } \exists \alpha \in \mathbb{Z} \ ext{ such that } 3600(k^2+1) = (\alpha)^2\n\Rightarrow \alpha = 60\sqrt{k^2 + 1}\n\Rightarrow \sqrt{k^2 + 1} \in \mathbb{Q} \Rightarrow k^2 + 1 \ ext{ is a perfect square}\n$$", "Wait — not necessarily. $ \sqrt{k^2 + 1} $ doesn’t have to be integer, but $ 3600(k^2 + 1) $ must be a square. Since 3600 is a perfect square, this occurs if and only if $ k^2 + 1 $ is a perfect square multiplied by a square factor — but actually, since 3600 is a square, $ k^2 + 1 $ must be a rational square, but since $ k \in \mathbb{Z} $, it follows that:", "Let $ k^2 + 1 = r^2 $, only if $ 3600(k^2 + 1) $ is a square whenever $ k^2 + 1 $ is a square times a square divisor — but more rigorously:", "Because 3600 is a square, $ 3600(k^2 + 1) $ is a square iff $ k^2 + 1 $ is a square times a square unit, but in integers, $ 3600(k^2 + 1) $ is a perfect square if and only if $ k^2 + 1 $ is a perfect square up to a square factor — actually, since 3600 carries the square, we just require:", "$$\nk^2 + 1 = \left( \frac{m}{60} \right)^2 \Rightarrow m^2 = 3600(k^2 + 1)\n$$", "So $ k^2 + 1 $ must be a square factor scaled appropriately — instead, let’s write:", "Let $ k^2 + 1 = s^2 $. Then $ 3600(k^2 + 1) = (60s)^2 $, a perfect square. But this assumes $ k^2 + 1 $ is a perfect square — which is too restrictive.", "Ah — correction: $ 3600(k^2 + 1) $ is a perfect square if and only if $ k^2 + 1 $ is a rational square. But since $ k \in \mathbb{Z} $, $ k^2 + 1 $ is a positive rational, and its square root is rational iff $ k^2 + 1 $ is a perfect square.", "Why? Because $ \sqrt{k^2 + 1} \in \mathbb{Q} $ only when $ k^2 + 1 $ is a perfect square (by rational root theorem or number theory).", "So:", "$$\n\exists \ ext{ integer } s \ ext{ such that } k^2 + 1 = s^2\n\Rightarrow s^2 - k^2 = 1\n\Rightarrow (s - k)(s + k) = 1\n$$", "This is a Pell-type equation. The only integer solutions are $ (s,k) = (\pm1, 0) $.", "But $ k = 0 \Rightarrow s^2 = 1 \Rightarrow s = \pm1 $, so $ y = 5k = 0 $", "Then:", "$$\n144(0)^2 + 3600 = 3600 = 60^2 \quad \ ext{perfect square}, \quad \ ext{and divisible by } 25\n$$", "But are there nonzero integers $ y $ such that $ 144y^2 + 3600 $ is a perfect square and divisible by 25?", "We revisit the condition:", "We found $ y \equiv 0 \pmod{5} $, so $ y = 5k $, leading to:", "$$\n144y^2 + 3600 = 3600(k^2 + 1) = m^2\n\Rightarrow \frac{m}{60} = \sqrt{k^2 + 1}\n\Rightarrow \ ext{must have } k^2 + 1 = \left( \frac{p}{q} \right)^2 \Rightarrow \ ext{over integers}, k^2 + 1 \ ext{ must be a rational square}\n$$", "But $ k^2 + 1 $ is an integer. $ \sqrt{k^2 + 1} \in \mathbb{Q} \Rightarrow \sqrt{k^2 + 1} = \frac{a}{b} $, reduced, implies $ k^2 + 1 = \frac{a^2}{b^2} \Rightarrow a^2 = b^2(k^2 + 1) $. Since $ k^2 + 1 \in \mathbb{Z} $, $ b^2 \mid a^2 $, so $ b $ divides $ a $. But for $ k^2 + 1 $ not a square, $ \sqrt{k^2 + 1} <br/>\notin \mathbb{Q} $, so $ 3600(k^2 + 1) $ is a square only if $ k^2 + 1 $ is a perfect square times a square factor, but more precisely:", "Let $ k^2 + 1 = d $. We want $ 3600d $ to be a perfect square.", "Since $ 3600 = 60^2 $, $ 3600d = 60^2 d $, so $ d $ must be such that $ 60^2 d $ is a perfect square ⇒ $ d $ must be a perfect square (because 3600 already has even exponents in prime factorization: $ 2^6 \cdot 3^2 \cdot 5^2 $, all exponents even). Therefore, $ 3600d $ is a perfect square iff $ d = k^2 + 1 $ is a perfect square.", "So $ k^2 + 1 = s^2 \Rightarrow s^2 - k^2 = 1 \Rightarrow (s - k)(s + k) = 1 $", "Integer solutions: $ s = \pm1, k = 0 $", "Thus, only solution is $ k = 0 $, so $ y = 0 $", "But wait — is that acceptable?", "Check: $ y = 0 \Rightarrow 144(0)^2 + 3600 = 3600 = 60^2 $, divisible by 25? $ 3600 \div 25 = 144 $, yes.", "But is there a nonzero $ y $?", "Suppose $ y <br/>\ne 0 $. Then $ k <br/>\ne 0 $, $ k^2 + 1 > 1 $. But $ k^2 + 1 $ is a positive integer greater than 1, not a perfect square (except $ k = 0 $), and since $ 3600(k^2 + 1) $ has $ (2^6 \cdot 3^2 \cdot 5^2)(k^2 + 1) $, the prime factorization gains odd exponent at primes dividing $ k^2 + 1 $ not in 3600. Since $ k^2 + 1 $ is an integer $ >1 $, unless it is a perfect square, $ 3600(k^2 + 1) $ is not a perfect square.", "Wait — but we only require $ 3600(k^2 + 1) $ to be a perfect square — not necessarily that $ k^2 + 1 $ is a square. But $ 3600(k^2 + 1) = (60)^2 (k^2 + 1) $. This is a perfect square iff $ k^2 + 1 $ is a perfect square — because 3600 has even exponents, so multiplying by $ k^2 + 1 $ introduces square-free part equal to that of $ k^2 + 1 $. Thus, the product is square iff $ k^2 + 1 $ is a perfect square.", "Hence, only solution is $ k^2 + 1 = s^2 \Rightarrow k = 0 $", "But $ y = 5k = 0 $", "Thus, the only integer $ y $ satisfying both conditions is $ y = 0 $", "Wait — let’s double-check divisibility: $ 144y^2 + 3600 \equiv 0 \pmod{25} $", "For $ y = 5 $: $ y^2 = 25 $, so:", "$ 144 \cdot 25 + 3600 = 3600 + 3600 = 7200 $", "$ 7200 \mod 25 = 0 $? $ 25 \ imes 288 = 7200 $, yes.", "$ 7200 = 60^2 \cdot 2 $? $ 60^2 = 3600 $, $ 7200 = 2 \cdot 3600 = 2 \cdot 60^2 $, not a perfect square.", "But $ 7200 = 60^2 \cdot 2 $, not a square.", "Try $ y = 5 \cdot 12 = 60 $: $ y^2 = 3600 $, $ 144 \cdot 3600 = 518400 $, $ 3600 + 518400 = 522000 $, $ \sqrt{522000} \approx 722.8 $, not integer.", "But recall: $ 144y^2 + 3600 = 3600(k^2 + 1) $, $ k = y/5 $", "Let $ y = 5k $, $ k \in \mathbb{Z} $. We need $ 3600(k^2 + 1) = m^2 $", "So $ m = 60\sqrt{k^2 + 1} $. For $ m $ integer, $ \sqrt{k^2 + 1} \in \mathbb{Q} \Rightarrow k^2 + 1 $ is a perfect square.", "As before, only solution $ k = 0 $", "But $ k^2 + 1 = s^2 \Rightarrow s^2 - k^2 = 1 \Rightarrow (s-k)(s+k)=1 \Rightarrow s = 1, k = 0 $ only integer solution.", "Thus, only solution is $ k = 0 \Rightarrow y = 0 $", "But wait — is $ 3600(k^2 + 1) $ ever a square for $ k <br/>\ne 0 $? For example, suppose $ k^2 + 1 = 2 $, then $ 3600 \cdot 2 = 7200 $, not square. $ k = 1 $: $ 2 \cdot 3600 = 7200 $. $ k = 2 $: $ 5 \cdot 3600 = 18000 $, not square. $ k = 3 $: $ 10 \cdot 3600 = 36000 $, not square. $ k = 4 $: $ 17 \cdot 3600 = 61200 $, not square.", "But what if $ k^2 + 1 = d^2 $? Only $ d = 1 $. So no other solutions.", "Alternatively, beyond Pell equations, suppose $ k^2 + 1 = 2m^2 $ — but that’s seating for larger equations, but we need $ k^2 + 1 $ to make $ 360"]